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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is parallel to the line , `2x+y=4` and the tangent to the ellipse at P passes through Q (4,4) then Pq is equal toA. `(5 sqrt(5))/(2)`B. `( sqrt(61))/(2)`C. `( sqrt(221))/(2)`D. `( sqrt(157))/(2)` |
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Answer» Correct Answer - A Key idea equation of tangent and normal to the `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at point `p(x_(V)Y_(1))` is ` T=0 =rarr (xx_(1))/(a^(2))+(y y_(1))/(b^(2))=1 and (a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)` respectively . Equation of given ellipse is `3x^(2)+4y^(2)=12` `implies (x^(2))/(4)+(y^(2))/(3)=1` Now . Let point `P(2 cos theta , sqrt(3) sin theta),` so equation of tangent to ellipse (i) at point P is `( x cos theta )/(2) +( y sin theta)/(sqrt(3))=1` since, tangent (ii) passes through point Q(4,4) ` therefore 2 cos theta +(4)/(sqrt(3))=1` and equation of normal to ellipse (i) nad point P is `(4x)/(2 cos theta)-(3y)/(sqrt(3)sin theta)=4-3` `implies 2x sin theta- sqrt(3) cos theta y = sin theta cos theta` since normal (iv) is parallel to line , 2x + y =4 `therefore ` Slope normal (iv) = slope of line `, 2x+y=4` `implies (2) /(sqrt(3)) tan theta =- 2implies tan theta =-sqrt(3) implies theta = 120^(@) ` ` implies ( sin theta , cos theta)=( (sqrt(3))/(2),-(1)/(2))` hence , point `p(-1,(3)/(2))` Now `PQ=sqrt((4+1)^(2)+(4-(3)/(2))^(2))` [ given cordinates of Q`-=`(4,4)] `=sqrt(25+(25)/(4))=(5sqrt(5))/(2)` |
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| 52. |
If a tangent of slope 2 of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`is normal to the circle `x^2+y^2+4x+1=0`, then the maximum value of `a b`is4 (b) 2(c) 1 (d)none of theseA. 4B. 2C. 1D. none of these |
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Answer» A tangent of slopw 2 to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is `y==4x+-sqrt(4a^(2)+b^(2))" "(1)` This is normal to the circle `x^(2)+y^(2)+4x+1=0` Therefore, (1) passes througj (-2,0), i.e., or `0=-4+-sqrt(4a^(2)+b^(2))` or `4a^(2)+b^(2)=16` Using `AMgeGM`, we get `(4a^(2)+b^(2))/(2)gesqrt(4a^(2)-b^(2))` or `able4` |
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| 53. |
If any tangent to the ellipse `(x^2)/(a^@)+(y^2)/(b^2)=1`interceptsequal lengths `l`on theaxes, then find `ldot` |
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Answer» Correct Answer - `sqrt(a^(2)+b^(2))` The equation of tangenet to the given ellipse at point `P(a cos theta, b sin theta)` is `(x)/(a) cos theta+(y)/(b) sin theta=1` The intercepts of line on the ellipse are `a//cos theta and b//sin theta`. Given that `(a)/(costheta)=(b)/(sin theta)l or cos theta=(theta)/(l)and sin theta=(b)/(l)` or `cos^(2)theta+sin^(2)theta=(a^(2))/(l^(2))+(b^(2))/(l^(2))` `or l^(2)=a^(2)+b^(2)` or `l=sqrt(a^(2)+b^(2))` |
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| 54. |
If the straight line `xcosalpha+ysinalpha=p`touches the curve `(x^2)/(a^2)+(y^2)/(b^2)=1`, then prove that `a^2cos^2alpha+b^2sin^2alpha=p^2dot` |
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Answer» We know that the line y=mc +c is a tangent to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` if `C^(2)=a^(2)m^(2)+b^(2)` Then comparing the given line `x cos theta+y sin alpha=p` with y=mx +c, we have c=`p//sin alpha,m=-cos alpha//sin alpha` So, the given line will be a tangent if `(p^(2))/(sin^(2)alpha)=a^(2)(cos^(2)alpha)/(sin^(2)alpha)+b^(2)` or `p^(2)=a^(2)cos^(2)alpha+b^(2)sin^(2)alpha)` |
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| 55. |
Find the slope of a common tangent to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`and aconcentric circle of radius `rdot` |
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Answer» Correct Answer - `sqrt((r^(2)-b^(2))/(a^(2)-r^(2)))` The equation of any tangent to the given ellipse is `y=mx+sqrt(a^(2)m^(2)+b^(2))` If it touches `x^(2)+y^(2)=r^(2)`, then `sqrt(a^(2)m^(2)+b^(2))=r sqrt(1+m^(2))` or `m^(2)(a^(2)-r^(2))=r^(2)-b^(2)` or `m=sqrt((r^(2)-b^(2))/(a^(2)-r^(2)))` |
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| 56. |
Extremities of the latera recta of the ellipses `(x^2)/(a^2)+(y^2)/(b^2)=1(a > b)` having a given major axis 2a lies onA. `x^(2) = a(a-y)`B. `x^(2) =a(a+y)`C. `y^(2) =a(a+x)`D. `y^(2) =a(a-x)` |
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Answer» Correct Answer - A::B Let (h,k) be the extremity of L.R. `:. h = +- ae, k = +- (b^(2))/(a)` `:. k= +-a (1-e^(2)) = +-a (1-(h^(2))/(a^(2))) = +- (a-(h^(2))/(a))` On taking + ve sign, we get `k = a- (h^(2))/(a)` `rArr (h^(2))/(a) = a -k rArr a-k rArr h^(2) =a (a-k)`. On taking -ve sign, we get `k =- a+(h^(2))/(a)` `rArr h^(2) = a(a+k)` |
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| 57. |
Let `E_1a n dE_2,`respectively, be two ellipses `(x^2)/(a^2)+y^2=1,a n dx^2+(y^2)/(a^2)=1`(where `a`is a parameter). Then the locus of the points of intersection of theellipses `E_1a n dE_2`is a set of curves comprisingtwo straight lines(b) one straight lineone circle(d) one parabolaA. two straigthsB. one straiths lineC. one circleD. one parabola |
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Answer» Let P(h,k) be the point of intersection of `E_(1) and E_(2)`. Then , `(h^(2))/(a^(2))+k^(2)=1` or `h^(2)=a^(2)(1-k^(2)) " "(1)` and `(h^(2))/(1)+(k^(2))/(a^(2))=1` `or k^(2)=a^(2)(1-h^(2))" "(2)` Eliminating from (1) and (2), we get `(h^(2))/(1-k^(2))=(k^(2))/(1-h^(2))` or `h^(2)(1-h^(2)=k^(2)(1-k^(2))` or `(h-k)(h+k)(h^(2)+k^(2)-1)=0` Hence, hte locus os a set of curves consisting of the straight lines y=x,y=-x, and the cirlce`x^(2)+y^(2)=1` |
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| 58. |
A chord is drawn passing through `P(2,2)` on the ellipse `(x^(2))/(25)+(y^(2))/(16) =1` such that it intersects the ellipse at A and B. Then maximum value of `PA.PB` isA. `(61)/(4)`B. `(59)/(4)`C. `(71)/(4)`D. `(63)/(4)` |
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Answer» Correct Answer - B Let equation of secant be `(x-2)/(cos theta) = (y-2)/(sin theta) =r` (Parametric form) Solving it with ellipse `((r cos theta+2)^(2))/(25)+((r sin theta +2)^(2))/(16)=1` `r^(2) (16 cos^(2) theta + 25 sin^(2) theta) +r (64 cos theta + 100 sin theta) - 236=0` `|r_(1)r_(2)| = PA. PB = |(-236)/(16 cos^(2)theta+25 sin^(2)theta)|` `=|-(236)/(16+9 sin^(2)theta)|` Max. `PA. PB = (236)/(16) = (59)/(4)` |
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| 59. |
A tangent having slope of `-4/3`to theellipse `(x^2)/(18)+(y^2)/(32)=1`intersectsthe major and minor axes at points `Aa n dB ,`respectively. If `C`is thecenter of the ellipse, then find area of triangle `A B Cdot` |
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Answer» Correct Answer - 24 sq. units For a given ellipse, the equation of tangent whsoe slope is m is `y=mx+sqrt(18m^(2)+32)` For m=-4/3, the tangent is `y=-(4)/(3)xxsqrt(18((16)/(9))+32)` `=-(4)/(3)x+18` or 4x+3y=24 This intersects the amjor and minor axes at (6,0) and B(0,8), respectively. |
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| 60. |
The auxiliary circle of a family of ellipsespasses through the origin and makes intercepts of 8 units and 6 units on thex and y-axis, respectively. If the eccentricity of all such ellipses is `1/2,`then find the locus of the focus. |
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Answer» The auxiliary circle makes intercepts of 8 units and 6 units on he x-and y-axes, respectively. Therefore, the centre of the circle is (4,3) and the radius is 10. (4,3) is the centre of inscirbed ellipse also. Now, the distance ofteh focus the from the centre of the ellipse is ae. Since eh radius of the auxiliary circle is 5, the length of the semimajor axis, a will be 5 . Given `e=(1)/(2)` `:. ae=(5)/(2)` Therefore, the locus of the focus is `(x-4)^(2)+(y-3)^(2)=(25)/(4)` |
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| 61. |
Prove that any point on the ellipse whose foci are`(-1,0)`and `(7,0)`andeccentricity is `1/2`is `(3+8costheta,4sqrt(3)sintheta),theta in Rdot` |
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Answer» The foci are (-1,0 ) and (7,0) . The distance between the foci is 2ae=8 or ae=4 Since e=1/2, we have a=8 Now, `b^(2)=a^(2)(1-e^(2))` or `b^(2)=48` or `b=4sqrt(3)`. The center of the ellipse is the midopoint of the joining two foci. Therefore, the coordinates of the center are (3,0) Hence its equation is `((x-3)^(2))/(8^(2))+((y-0)^(2))/((4sqrt(3)))=1 " "(1)` Thus, hte parameter coordinates of a point on (1) are `(3+8 cos theta, 4 sqrt(3)sin theta)` |
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| 62. |
Find the equation of the curve whose parametricequation are `x=1+4costheta,y=2+3sintheta,theta in Rdot` |
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Answer» We have, `x=1+4 cos theta, y=2+3 sin theta`. Therefore , `((x-1)/(4))^(2)+((y-2)/(3))^(2)=1` or `((x-1))/(16)+((y-2)^(2))/(9)=1` which is an ellipse. |
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| 63. |
How many tangents to the circle `x^2 + y^2 = 3` are normal tothe ellipse `x^2/9+y^2/4=1?`A. 3B. 2C. 1D. 0 |
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Answer» Correct Answer - D Equation of normal to ellipse `(x^(2))/(9) + (y^(2))/(4) =1` at `P(3 cos theta, 2 sin theta)` is `3x sec theta -2y cos theta =5` If it is tangent to circle `x^(2) + y^(2) =3`, then `rArr (5)/(sqrt(9sec^(2) theta+4 cosec^(2)theta)) =sqrt(3)` `9 sec^(2) theta + 4 cosec^(2) theta = 9 = 4+ 9 tan^(2) theta + 4 cot^(2) theta = 25 +(3 tan theta -2 cot theta)^(2)` `:. (9 sec^(2) theta + 4 cosec^(2) theta)_(min) =25` `:.` no such `theta` exists. Hence no tangents to circle which is normal to ellipse. |
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| 64. |
Match the following lists: |
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Answer» (a) The equation of tangent at `((cos theta)//2,(sin theta)//3) "is" 2x cos theta+3ysin theta=1` Which is parallel to the given line `8x=9y`. Therefore, `cos theta=+-(4)/(5), sin =(+-3)/(5)` Hence, the points are (2/5,-1/5) and (-2/51/5). The distance between the points is `sqrt((16)/(25)+(4)/(25))=(2)/(sqrt(5))` Which is the given The given equation is `((x+1)^(2))/(9)+((y+2)^(2))/(25)=1` or `e^(2)=1-(9)/(25)=(16)/(25)or e=(4)/(5)` Hence, the foci, `S,S-=(1,-2+-4)-=S(-1,2) and S(-1.-6)` Required sum of distances `=2+6=8` (c) The equation of normal at `(3sos theta, 2 sin theta)` s `3xsec theta-2y "cosec" theta=5` whis is parallel to the given line `2x+y=1`. Therefore, `cos theta=+-(3)(5), sin theta=+-(4)/(5)` Hence, the points are `(-9/5,9/5) and (9/5,-8/5)` Required sum of the distances`=(16)/(5)` (d) Conider any point `(t,t+2), t in R` on the line `x-y+2=0` The chord of contact of the ellipse w.r.t. this point is `xt+2y(t+2)=2` or `(4y-2)+(x+2y)=0` This line passes trough the points of the intersection of lines `4y-2=0 and x+2y=0` `:. x=-1,y=1//2` Hence, the point is `(-1,1//2)` os 3. |
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| 65. |
Match the following lists: |
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Answer» (a) Tangent to the ellipe at `P(phi)` is `(x)/(4) cos phi+(y)/(2) sin phi=1` It must pass through the center of the circle. Hence, `(4)/(4) cos phi+(2)/(2)sin phi=1` `or cos phi+sin phi=1` or `1+sin 2 phi phi=1` or `sin 2phi=0` i.e.,` 2phi=0 or pi` i.e., `(phi)/(2)=0 or (pi)/(4)` (b) Consider and point `P(sqrt(6) cos, theta, sqrt(2) sin theta)` on the ellipe `(x^(2))/(6)+(y^(2))/(2)=1` Given thast P=2. Therefore, `6 cos^(2) theta+2sin^(2) theta=0` or `4cos^(2) theta=2` or `cos theta=+-(1)/(sqrt(2))` i.e.,` theta=(pi)/(4) or (5pi)/(4)` (c) Sovling the equation of ellipse and parbola (eliminating `(x^(2)`)n we have `y-1+4y^(2)=4` or `4y^(2)+y-5=0` `(4y+5)(y-1)=0` or y=1, x=0` The curves touch at (0,1) so, the angle of intersection is 0. (d) The normal at`P(a cos theta, b sin theta)` is `(ax)/(cos thet)-(hx)/(sin theta)=a^(2)-b^(2)` Where `a^(2)=14,b^(2)=5` It means the cruve again at Q `(2 theta)`, i.e., `(a cos 2 theta, b sin 2theta)` Hecen, `(a)/(costheta)a cos 2theta-(b)/(sin theta)(b 2sintheta)=a^(2)-b^(2)` `or (14)/(cos theta) cos 2theta-(5)/(sin theta)(sin 2 theta)=14-5` or `28 cos^(2)theta-14-10 sec^(2)theta=9 cos theta` `or 18 cos^(2)theta-9cos theta-14=0` `or (6 cos theta-7)(3 cos theta+2)` or `cos theta=-(2)/(3)` |
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| 66. |
Find the equation of an ellipse whose eccentricity is 2/3, the latusrectum is 5 and the centre is at the origin. |
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Answer» Correct Answer - `(4x^(2))/81 + (4y^(2))/45 = 1` `(2b^(2))/a = 5 rArr (2a^(2)(1-e^(2)))/a = 5 rArr a = 9/2.` Also, ` b^(2) = (5a)/2 = (5/2 xx 9/2) = 45/4.` |
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| 67. |
Find the eccentricity of the ellipse whose (i) latus rectum is half of minor axis (ii) minor axis is half of major axis. |
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Answer» Correct Answer - ` e = sqrt3/2 ` `{(2b^(2))/a = 1/2 xx 2b rArr b = a/2}.` So, ` b^(2) = a^(2)(1-e^(2)) rArr a^(2)/4 = a^(2) (1-e^(2)).` |
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| 68. |
Find the equation of the ellipse, whose length of the major axis is 20 and foci are `(0,+-5)`. |
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Answer» Since the foci of the ellipse lie on the y-axis, it is a vertical ellipse. Let the required equation be ` x^(2)/b^(2) + y^(2)/a^(2) = 1," where " a^(2) gt b^(2).` Let ` c^(2) = (a^(2) -b^(2)).` Its foci are `(0, pm c) ` and therefore, c = 5. Also, a = length of the semi-major axis = `(1/2 xx 20) = 10.` Now, `c^(2) = (a^(2) -b^(2)) hArr b^(2) = (a^(2)-c^(2)) = (100 - 25) = 75`. Thus, `a^(2) = (10)^(2) = 100 and b^(2) = 75.` Hence, the required equation is `x^(2)/75 + y^(2)/100 = 1.` |
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| 69. |
Find the equation of an ellipse whose vertices are `(0,+-10)`and eccentricity `e=4/5` |
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Answer» Since the vertices of the ellipse lie on the y-axis, it is a vertical ellipse. Let the required equation be `x^(2)/b^(2) + y^(2)/a^(2) = 1," where " a^(2) gt b^(2).` Its vertices are `(0, pm a) ` and therefore, a = 10. Let ` c^(2) = (a^(2) - b^(2)).` Then , ` e = c/a rArr x = ae = (10 xx 4/5) = 8`. Now, `c^(2) = (a^(2)-b^(2)) hArr b^(2) = (a^(2)-c^(2)) = (100 - 64) = 36.` `:. a^(2) = (10)^(2) = 100 and b^(2) = 36.` Hence, the required equation is ` x^(2) / 36 + y^(2)/ 100 = 1. ` |
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| 70. |
Variable complex number z satisfies the equation `|z-1+2i|+|z+3-i|=10`. Prove that locus of complex number z is ellipse. Also, find the centre, foci and eccentricity of the ellipse. |
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Answer» Given equation is `|z-(1-2i)|+|z-(-3+i)|=10` `|z-(1-+i)|`= distacne between z and 1-2i`(=z+(1))` `|z-(-3_(i))|`= distacne between z and-+i`(=z_(2))` Thus, sum of distance of z form points (1-2i) and (-3+i) is constant 10. So, locus of z is ellips e with foci at `z_(1) and z_(2)`. Centre is midpoint of `z_(1) and z_(2)`, which is `-1-(i)/(2)` Distane between foci = `|z_(1)-z_(2)|-|1-2i-(-3+i)|=|4-3i|=5 = 2ae` Major axis =10=2a `:.` Eceentricity `=e=(5)/(10)=(1)/(2)` |
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| 71. |
Find the equation of ellipse, Centre at (0,0), major axis on the y-axis and passes through the points (3,2) and (1,6). |
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Answer» Since the major axis of the ellipse lies on the y-axis, it is a vertical ellipse. Let the required equation be `x^(2)/b^(2) + y^(2)/a^(2) = 1" "("where " a^(2) gt b^(2))." "` …(i) Since (3, 2) lies on (i) , we have ` 9/b^(2) + 4/a^(2) = 1." "` ...(ii) Also, since (1, 6) lies on(i), we have `1/b^(2) + 36/a^(2) = 1." " ` ...(iii) Putting ` 1/b^(2) = u and 1/a^(2) = v`, these equations become: ` 9u+4v = 1 " "` ...(iv) ` and u + 36 v = 1" " ` (v) On multiplying (v) by 0 and subtracting (iv) from it, we get ` 320v = 8 hArr v = 8/(320) = 1/(40) hArr 1/a^(2) = 1/40 hArr a^(2) = 40.` Putting ` v = 1/40 ` in (v), we get ` u + (36 xx 1/40) = 1 hArr u = (1-9/10) = 1/10 hArr 1/b^(2) = 1/10 hArr b^(2) = 10.` Thus, ` b^(2) and a^(2) = 40.` Hence, the required equation is ` x^(2)/10 + y^(2) / 40 = 1.` |
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| 72. |
Find the equation for the ellipse that satisfies the givenconditions:Major axis on the xaxis and passes through the points (4, 3) and(6, 2). |
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Answer» Since the major axis of the ellipse lies on the x-axis, so it is a horizontal ellipse. Let the required equation of the ellipse be `x^(2)/a^(2) +y^(2)/b^(2) = 1" "("where "a^(2) gt b^(2))." "` …(i) Since (4, 3) lies on (i), we have `16/a^(2) +9/b^(2) = 1." "` ...(ii) Also, since (6, 2) lies on (i) , we have `36/a^(2) + 4/b^(2) = 1." "`...(iii) Putting ` 1/a^(2) = u and 1/b^(2) = v ` in (ii) and (iii), we get `16u+9v=1" "`...(iv) `36u+4v=1" " ` ...(v) On multiplying (iv) by 9 and (v) by 4, and subtracting, we get ` 65v = 5 hArr v = 1/13 hArr 1/b^(2) = 1/13 hArr b^(2) = 13.` Putting ` v = 1/13 ` in (iv), we get ` 16 u = (1-9/13) hArr 16u = 4/13 hArr u = (4/13 xx 1/16) = 1/52 hArr 1/a^(2) = 1/52 hArr a^(2) = 52.` Thus, `a^(2) 52 and b^(2) = 13.` Hence, the required equation is `x^(2)/52 + y^(2)/13 = 1.` |
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| 73. |
Comprehension- I A coplanar beam of light emerging from a point source have equation `lambda x-y+2(1+lambda)=0,lambda in R.` The rays of the beam strike an elliptical surface and get reflected. The reflected rays form another convergent beam having equation `mu x-y+2(1-mu)=0,mu in R.` Foot of the perpendicular from the point (2, 2) upon any tangent to the ellipse lies on the circle `x^2 + y^2 - 4y - 5 = 0` The eccentricity of the ellipse is equal toA. `1//3`B. `1//sqrt(3)`C. `2//3`D. none of these |
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Answer» `lambdax-y+2(1+lambda)=0` or `lambda(x+2)-(y-2)=0,lambda in R` This line passes through `(-2,2)` `mux-y+2(1-mu)=0, mu in R` or `mu(x-2)-(y-2)=0` This line passes through (2,2) Clearly, (-2,2) and (2,2) reprsente the foci of the ellipse So, 2ae=4 The circle `x^(2)+y^(2)-4y-5=0i.e.,x^(2)+(y-2)^(2)=9` respresent an auxiliar circle. Thus, `a^(2)=9 or e=2//3 and b^(2)=5` |
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| 74. |
A coplanar beam of ligth emerging from a point source has the equation `lambdax-y+2(1+alambda)-0, lambda in R.` The rays of the beam strike an elliptical surface and get reflected. The reflected rays form another convergent beam having eqution `mux-y+2(1-mu)=0, mu in R`. Further, it is found that the foot of the perpendicular from the point (2,2) upon any tangent to the ellipse lies on the circle `x^(2)+y^(2)-4y-5=0` The total distance travelled by an incident ray starting at one focus and the correcponding reflected ray terminating at another focus isA. 9B. 8C. 7D. 6 |
| Answer» Required area `=(1)/(2)(2ae)(b)=3xx(2)/(3)xxsqrt(5)=2sqrt(5)` | |
| 75. |
A coplanar beam of ligth emerging from a point source has the equation `lambdax-y+2(1+alambda)-0, lambda in R.` The rays of the beam strike an elliptical surface and get reflected. The reflected rays form another convergent beam having eqution `mux-y+2(1-mu)=0, mu in R`. Further, it is found that the foot of the perpendicular from the point (2,2) upon any tangent to the ellipse lies on the circle `x^(2)+y^(2)-4y-5=0` The area of the largest triangle that na incident ray and the corresponding reflected rat can enclose with the axis of the ellipse is equal toA. `4sqrt(5)`B. `2sqrt(5)~C. `sqrt(5)`D. none of these |
| Answer» Required area `=(1)/(2)(2ae)(b)=3xx(2)/(3)xxsqrt(5)=2sqrt(5)` | |
| 76. |
The slopes of the common tanents of the ellipse `(x^2)/4+(y^2)/1=1`and the circle `x^2+y^2=3`are`+-1`(b) `+-sqrt(2)`(c) `+-sqrt(3)`(d) none of theseA. `+-1`B. `+-sqrt(2)`C. `+-sqrt(3)`D. none of these |
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Answer» Let m be the slop of the common tanent .Then, `+-sqrt(3)sqrt(1+m^(2))=+-sqrt(4m^(2)+1)` or `3+2m^(2)=4m^(2)+1` or `m^(2)=2` or `m=+-sqrt(2)` |
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| 77. |
If the line y=mx+c is a tangent to the ellipse `x^(2)+2y^(2)=4`,A. `-sqrt(2)`B. `sqrt(2)`C. 2D. 1 |
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Answer» Using condition of tangency, we get `c^(2)=4m^(2)+2ge2AA m in R` `:.c^(2)ge2` `rArr|c|gesqrt(3)` |
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| 78. |
Find the normal to the ellipse `(x^2)/(18)+(y^2)/8=1`at point (3, 2). |
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Answer» Equation of normally to ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at point `P(x_(1),y_(1))` is `(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)` So, equationn of normally to ellipse `(x^(2))/(18)+(y^(2))/(8)=1` at point P(3,2) `(18x)/(3)-(8y)/(2)j=18-8` or 3x-2y=5 |
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| 79. |
the value of `lambda` for which the line `2x-8/3lambday=-3` is a normal to the conic `x^2+y^2/4=1` is: |
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Answer» Given normal to ellipse is `2x-(8)/(3)lamday=-3` or `y=((3)/(4lambda))x+((9)/(8lambda))` Here, `m=(3)/(4lambda)=(9)/(8lambda)` Now, condition for line y=mx+c to be normed to ellipse is `C=+-((a^(2)-b^(2))m)/(sqrt(a^(2)+b^(2)m^(2)))` `rArr(9)/(8lambda)=+-(3(3)/(4lambda))/(sqrt(1+4(9)/(16lambda^(2))))` `rArr 4lambda^(2)+9=16lambda^(2)` `rArr4lambda^(2)=3` `rArr=+-(sqrt(3))/(2)` Thus, normals are `2x+-(4)/(sqrt(3))y=-3` |
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| 80. |
Find the points on the ellipse `(x^2)/4+(y^2)/9=1`on whichthe normals are parallel to the line `2x-y=1.` |
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Answer» The normal at `P(2 cos theta,3 sin theta)` is `(2x)/(cos theta)-(3y)/(sin theta)=-5` Now, this normal is parallel to 2x-y1. Then, `(2//cos theta)/(3//sin theta)=2` or ` tan theta=(3)/(1)` `cos theta=+-and sin theta=+-(3)/(sqrt(10))` Hence, the points are `(2sqrt(10),9sqrt10),(-2//sqrt(10),-9//sqrt(10))` |
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| 81. |
Match the conics in List I with the statements/expressions in List II |
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Answer» `x=sqrt(3)((1-t^(2))/(1+t^(2))),y=(2t)/(1+t^(2))` Let `t= tan alpha` `:.cos 2alpha=(x)/(sqrt(3)) and sin 2alpha=y` `:. (x^(2))/(3)+y^(2)=sin^(2)2alpha+cos^(2)2alpha=1` Note : Solutions of the remainnig parts are given in their respective chapters. |
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| 82. |
a triangle `A B C`with fixed base `B C`, the vertex `A`moves such that `cosB+cosC=4sin^2A/2dot`If `a ,ba n dc ,`denote the length of the sides of the triangle opposite to the angles `A , B ,a n dC`, respectively, then`b+c=4a`(b) `b+c=2a`the locus of point `A`is an ellipsethe locus of point `A`is a pair of straight linesA. b+c=4aB. b+c=2aC. the locus of point A is an ellipseD. the locus of point A is a pair of straight lines |
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Answer» Given `cos B+cos C=4 sin^(2).(A)/(2)` `or 2 cos((B+C)/(2))cos ((B-C)/(2))=4 sin^(2).(A)/(2)` or `cos.((B-C)/(2))=2sin.((A)/(2))` `or 2 sin.(B+C)/(2)cos.(B-C)/(2)=4 sin.(A)/(2)cos.(A)/(2)` or `sin B+sin C=2 sin A` or `b+c=2a` Thus, sum of two variable sides is constant . Hence the locus of vertex A is an ellipse with B and C as foci, |
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| 83. |
If the tangent drawn at point `(t^2,2t)`on the parabola `y^2=4x`is the same as the normal drawn at point `(sqrt(5)costheta,2sintheta)`on the ellipse `4x^2+5y^2=20,`then`theta=cos^(-1)(-1/(sqrt(5)))`(b) `theta=cos^(-1)(1/(sqrt(5)))``t=-2/(sqrt(5))`(d) `t=-1/(sqrt(5))`A. `theta=cos^(-1)(-(1)/(sqrt(5)))`B. `theta=cos^(-1)((1)/(sqrt(5)))`C. `t=-(2)/(sqrt(5))`D. `t=-(1)/(sqrt(5))` |
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Answer» The eqaution of the tagent at `(t^(2),2t)` to the parabola `y^(2)=4x` is `2ty=2(x+r^(2))` or`ty=x+r^(2)` or `x-ty+t^(2)=0" "(1)` The equation of the normal at `(sqrt(5)cos theta, 2 sin theta)` on the ellipse `4x^(2)+5y^(2)=20`is ` (sqrt(5)sectheta)x-(2"cosec" theta)y=5-4` ` (sqrt(5)sectheta)x-(2"cosec" theta)y=1" "(2)` Given that (1) and (2) represent the same line then. `(sqrt(5)sectheta)/(1)=(-2"cosec" theta)/(-t)=(-1)/(r^(2))` or `t=(2)/(sqrt(5))cot theta and t=-(1)/(2)sintheta` or `(2)/(sqrt(5))cot theta =-(1)/(2)sintheta` or `4 cos theta=-sqrt(5)(1-cos^(2)theta)` on `sqrt(5)cos^(2) theta-4 costheta-sqrt(5)=0` or `(cos theta-sqrt(5))(sqrt(5)cos theta+1)=0` `or cos theta=-(1)/(sqrt(3))` `or theta=cos^(-1)(-(1)/(sqrt(5)))` Putting `cos theta-ssqrt(5) "in" t=-(1//2) sin theta,` we get `t=-(1)/(2)sqrt(1-(1)/(5))=-(1)/(sqrt(5))` Hence , `theta = cos^(-1)(-(1)/(sqrt(5))) and t=-(1)/(sqrt(5))` |
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| 84. |
Let `E_(1) and E_(2)` be two ellipse whsoe centers are at the origin. The major axes of `E_(1) and E_(2)` lie along the x-axis , and the y-axis, respectively. Let S be the circle `x^(2)+(y-1)^(2)=2` . The straigth line x+y=3 touches the curves, S, `E_(1) and E_(2)` at P,Q and R, respectively . Suppose that `PQ=PR=(2sqrt(2))/(3)`. If `e_(1) and e_(2)` are the eccentricities of `E_(1) and E_(2)` respectively, thent hecorrect expression (s) is (are)A. `e_(1)^(2)+e_(2)^(2)=(43)/(40)`B. `e_(1)e_(2)=(sqrt(7))/(2sqrt(10))`C. `|e_(1)^(2)-e_(2)^(2)|=(5)/(8)`D. `e_(1)e_(2)=(sqrt(3))/(7)` |
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Answer» Let ellipse be `E_(1):(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and E_(2)=(x^(2))/(A^(2))+(y^(2))/(B^(2))=1` Sincex+y-3 is a tangent, `a^(2)+b^(2)=A^(2)+b^(2)=9` (using condition `c^(2)=a^(2)m^(2)+b^(2)` etc). Point P lies on `x^(2)+(y-1)^(2)=2` Euation of normal to circle having slop 1 is `y-1=1x(x-0) or y-y+1=0` Solving this normal with tangent line we get point P(1,2) Now, `PA=PR=(2sqrt(2))/(3)` So, points on line `x+y-3=0` at distance `(2sqrt(2))/(3)` from point `P "are"(1+-(1)/(sqrt(2))(2sqrt(2))/(3),2+-(1)/(sqrt2)(2sqrt(2))/(3))` or `Q((5)/(3),(4)/(3)) and Q ((1)/(3),(8)/(3))` Now, `Q((5)/(3),(4)/(3))` lies on `E_(1)` So, `(25)/(9a^(2))+(16)/(9(9-a^(2)))=1` `rArr2525-25a^(2)+16a^(2)=9a^(2)(9-a^(2))` `rArra^(2)-10a^(2)+25=0` `rArr a^(2)=5 so b^(2)=4` `:. e_(1)^(2)=1(b^(2))/(a^(2))=1-(4)/(5)=(1)/(5)` Now, `((1)/(3),(8)/(3))` lies on `E_(2)` So, `(1)/(A^(2))+(64)/((9-A^(2))=9` `rArr9-A^(2)+64A^(2)=9A^(2)(9-A^(2))` `rArr-2A^(2)+1=0` `rArr A^(2)=1 " " so, B^(2)=8` `:. =1- =(1)/(8)=(7)/(8)` |
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| 85. |
If from a point `P ,`tangents `P Qa n dP R`are drawnto the ellipse `(x^2)/2+y^2=1`so that theequation of `Q R`is `x+3y=1,`then findthe coordinates of `Pdot` |
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Answer» Let coordinates of P be `(x_(1),y_(1))` Then equation of QR is ltbr?`(x_(1)x)/(2)+y_(1)y)=1" "` ltbr But given equation of QR is `x+3y=1" "(2)` Equation (1) and (2) must be identical. `:. (x_(1))/(2)=(y_(1))/(3)=1` `rArrx_(1)=2 and y_(1)=3` `:. P-=(2,3)` |
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| 86. |
If there exists exactly one point of the line `3x+4y+25=0` from which perpendicular tangesnt can be brawn to ellipse `(x^(2))/(a^(2))+y^(2)=1(agt1)`, |
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Answer» Locus of point of intersecrtion fo perpendicular tangents is director circle. If there axists exactly one such point on the line 3x+4y+25=0, then line must touch the dierector circle whose equation is `x^(2)+y^(2)=a^(2)=1` Distance of centre of the circle from the line is 5. So, we must have `25=a^(2)+1` `rArr a^(2)=24 :. a =sqrt(24)` `:.` Electricity `sqrt(1-(1)/(24))=sqrt((23)/(24))` |
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| 87. |
An ellipse slides between two perpendicular lines the locus of its centre , is |
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Answer» Ellipse slides between two perpendicular lines. So, these lines are perpendicular tangents and their point of intersection P lies on the director circle. If centre of the ellipse is fixed, then all the points of intersection of perpendicular tangents lie at at fixed distance which is equal to radius of the director circle. So, if point of intersection of perpendicular tangents is fixed (point P,) then centre of variable ellipe also lies at fixed distance from point P. Therefore, locus of centre of the ellipse is a circle. |
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| 88. |
Find the angle between the pair of tangents fromthe point (1,2) to the ellipse `3x^2+2y^2=5.` |
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Answer» The combined eqution of the pair of tangents drawn from (1,2) ot the ellipse `3x^(2)+2y^(2)=5` is `(3x+4y-5)^(2)=(3x^(2)+3y^(2)-5)(3+8-5)[ "Using" T^(2)=SS_(1)]` `or 9x^(2)-24xy-4y^(2)+.....=0` If angle between pair of lines is `theta`, then `tan theta=(2sqrt(h^(2)-ab))/(a+b)` where a=9,h12, b=-4 `:. tan theta =(12)/(sqrt(5))` `rArr theta = tan^(-1).(12)/(sqrt(5))` |
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| 89. |
The muinimum area of the triangle formed by the tangent to `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and the coordinate axes isA. ab sq. unitsB. `(a^(2)+b^(2))/(2)` sq. unitsC. `((a+b)^(2))/(2)` sq. unitsD. `(a^(2)+ab+b^(2))/(3)` sq. units |
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Answer» Tangent to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at `p( a cos theta, b sin theta)` is given by `(xcos theta)/(a)+(y sin theta)/(b)=1` It meets he coordinate axes at `A(a sec theta, 0) and B(0, b "coses" beta)` or `Delta=(ab)/(sin 2 theta)` For area to be minimum, `sin theta` should be maxiumum and we know that teh maxiumum value of `sin theta` is 1. Therefore, `Delta_("max")=ab` |
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| 90. |
Find the equation of chord of an ellipse `(x^(2))/(25)+(y^(2))/(16)=1` joining two points `P((pi)/(3))and Q ((pi)/(6))` |
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Answer» Equation of chord PQ is `(x)/(5)cos(((pi)/(6)+(pi)/(3))/2)+(y)/(4)sin(((pi)/(6)+(pi)/(3))/2)=cos(((pi)/(6)-(pi)/(3))/2)` `rArr(x)/(5)cos.(pi)/(3)4+(y)/(4)sin.(pi)/(4)=cos.(pi)/(12)` `rArr(x)/(5)(1)/(sqrt(2))+(y)/(4)(1)/(sqrt(2))=(sqrt(3)+1)/(2sqrt(2))` `rArr(x)/(5)+(y)/(4)=(sqrt(3)+1)/(2)` |
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| 91. |
The eccentric angle of a point on the ellipse `(x^2)/4+(y^2)/3=1`at a distance of 5/4 units from the focus on the positive x-axis is`cos^(-1)(3/4)`(b) `pi-cos^(-1)(3/4)``pi+cos^(-1)(3/4)`(d) none of theseA. `cos^(-1)(3//4)`B. `cos^(-1)(4//5)`C. `cos^(-1)(3//5)`D. none of these |
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Answer» Any point on the ellipse is `(2cos theta, sqrt(3) sin theta)` The focus on the positive x-axis is (1,0) Given that `(2 cos theta-1)^(2)+3sin^(2)theta=(25)/(16)` or `cos theta=(3)/(4)` |
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| 92. |
The eccentricity of the locus of point `(2h+2,k),`where `(h , k)`lies on the circle `x^2+y^2=1`, is`1/3`(b) `(sqrt(2))/3`(c) `(2sqrt(2))/3`(d) `1/(sqrt(3))`A. `1//3`B. `sqrt(2)//3`C. `2sqrt(2)//3`D. `1//sqrt(3)` |
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Answer» (1) Let p =3h+2 and q=k or `h=(p-2)/(3) and k=q` Since (h,k) lies on `x^(2)+y^(2)=1` `h^(2)+k^(2)=1` or `((p-2)/(3))^(2)+q^(2)=1` The locus is (2) `((x-2)/(3))^(2)+y^(2)=1` which has eccentricity `e=sqrt(1-(1)/(9))=(2sqrt(2))/(3)` |
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| 93. |
If the ellipse `(x^2)/4+y^2=1`meets the ellipse `x^2+(y^2)/(a^2)=1`at four distinct points and `a=b^2-5b+7,`then `b`does not lie in`[4,5]`(b) `(-oo,2)uu(3,oo)``(-oo,0)`(d) `[2,3]`A. [4,5]B. `(-oo,2)uu(3,oo)`C. `(-oo,0)`D. [2,3] |
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Answer» For the two ellipse to intersect a four distinct point `agt1` `:. b^(2)-5b+7gt` `or b^(2)-5b+6gt0` or `b in (-oo,2)uu(3,oo)` Therefore, b does not lie in [2,3] |
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| 94. |
Tangents are drawn from the point P(3, 4) to the ellipse `x^2/9+y^2/4=1` touching the ellipse at points A and B.A. `9x^(2)+y^(2)-6xy- 54 x- 62y+241=0`B. `x^(2) +9x^(2)+6xy-54 x+62y- 241=0`C. `9x^(2) +9y^(2)+6xy-54 x-62y- 241=0`D. `x^(2)+y^(2)-2xy+27 +31y-120=0` |
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Answer» Correct Answer - A Equation of `AB" is" y-0 =-(1)/(3)(x-3)` `x+3y-3=0` `implies |x+3y-3|^(2)=10[(x-3)^(2)+(y-4)^(2)]` on solving , we are getting `9x^(2)+y^(2)-6xy- 54x-62y+241=0` |
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| 95. |
A bar of given length moves with its extremitieson two fixed straight lines at right angles. Show that any point on the bardescribes an ellipse. |
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Answer» OX and OY are fixed lines AB is bar, Let point P(x,y) on AB it divides the line AB in a:b draw PM`_|_`OY and PL`_|_`OX `PM=x=acostheta` `x/a=costheta` `PL=y=bsintheta` `y/b=sintheta` `cos^2theta+sin^2theta=1` `(x/a)^2+(y/b)^2=1` `x^2/a^2+y^2/b^2=1`. |
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| 96. |
Find the eccentricity, centre, vertices, foci,minor axis, major axis, directrices and latus-rectum of theellipse `25 x^2+9y^2-150 x-90 y+225=0.` |
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Answer» `x^2/a^2+y^2/b^2=1` `25x^2+9y^2-150x-90y+225=0` `25(x^2-6x)+9(y^2-10y)+225=0` `25(x^2-6x+9-9)+9(y^2-10y+25-25)+225=0` `25(x-3)^2-225+9(y-5)^2-225+225=0` `(25(x-3)^2)/225+(9(y-5))^2/225=1` `(x-3)^2/9+(y-5)^2/25=1` `x=X+3` `y=Y+5` `[x^2/3^2+Y^2/5^2=1]` `e=sqrt(1-a^2/b^2)=sqrt(1-9/25)=4/5` `x=0+3=3` `y=Y+5=Y+5=5+5=10` `y-5=5-5=0` `(3,0),(3,10)` `X=0,Y=pmbe` `x=X+3` `y=Y+5` `=-4+5=1` Equation of direction X and Y `Y=pmb/e=pm5/(4/6)=pm25/4` `y=Y+5=25/4+5=45/4` `y=445/4,y=-5/4` Major axis=`2b=2*5=10` Minor axis=`2a=2*3=6` Latus rectum=`(2a^2)/b=(2*9)/5=18/5` Length of latus rectum. |
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| 97. |
The ratio of the area of triangle inscribed inellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`to that of triangle formed by the corresponding points onthe auxiliary circle is 0.5. Then, find the eccentricity of the ellipse. |
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Answer» The given ratio is `(b)/(a)=(1)/(2)` Now, `e^(2)=1-(b^(2))/(a^(2))=1-(1)/(4)=(3)/(4)or e=(sqrt(3))/(2)` |
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| 98. |
If a triangle is inscribed in a n ellipse andtwo of its sides are parallel to the given straight lines, then prove thatthe third side touches the fixed ellipse. |
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Answer» Let the eccntric angls of the vertices P,Q, R of `DeltaPQR "be" theta_(1),theta_(2),theta_(3)`, respectively. Then the equations of PQ and PR are, respectively, `(x)/(a)cos.(theta_(1)+theta_(2))/(2)+(y)/(b) sin.(theta_(1)+theta_(2))/(2)=cos.(theta_(1)-theta_(2))/(2)` `and (x)/(a)cos.(theta_(2)+theta_(3))/(2)+(y)/(b) sin.(theta_(2)+theta_(3))/(2)=cos.(theta_(2)-theta_(3))/(2)` If PQ and PR are parallel to the given straigth lines, then we have `theta_(1)+theta_(2)` (Constatn = `2alpha` (say) and `theta_(1)+theta_(3)` = Constant= `2beta" "(1)` Hence, `theta_(1)-theta_(3)= 2 (alpha-beta)` Now, the equation of QR is `(x)/(a)cos.(theta_(2)+theta_(3))/(2)+(y)/(b)sin.(theta_(2)+theta_(3))/(2)=cos.(theta_(2)-theta_(3))/(2)" "(2)` or `(x)/(a)cos.(theta_(2)+theta_(3))/(2)+(y)/(b)sin.(theta_(2)+theta_(3))/(2)=cos(alpha-beta)" "(3)` Which shows that line (2), for different values of `theta_(2)+theta_(3)` , is tangent to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=cos^(2)(alpha-beta)` |
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| 99. |
The number of values of `c`such that the straight line `y=4x+c`touches the curve `(x^2)/4+(y^2)/1=1`is0 (b) 1(c) 2 (d) infinite |
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Answer» Correct Answer - B for ellipse , condition of tangency is `c^(2)= a^(2) m^(2) +b^(2)` given line is y= 4x+ c and curve `(x^(2))/(4)+y^(2)=1` `implies c^(2) = 4xx4^(2)+1=65` `implies c=+- sqrt(64)` so , there are two different values of C. |
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| 100. |
An ellipse is drawn by taking a diameter of the circle `(x)^2+""y^2=""1`as its semiminor axis and adiameter of the circle `x^2+""(y""""2)^2=""4`as its semi-major axis. If thecentre of the ellipse is the origin and its axes are the coordinate axes,then the equation of the ellipse is(1) `4x^2+""y^2=""4`(2) `x^2+""4y^2=""8`(3) `4x^2+""y^2=""8`(4) `x^2+""4y^2=""16` |
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Answer» `(x-1)^2 + y^2 = 1` `r=1` `d=2` `b=2` and `x^2 + (y-2)^2 = 4` `r=2` `d= 4` `a=4` eqn of ellipse `x^2/a^2 + y^2/b^2 = 1` `x^2/16 + y^2/4 = 1` `(x^2 + 4y^2)/16=1` `x^2 + 4y^2 = 16` option 3 is correct |
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