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InterviewSolution
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.
| 1. |
The number of points on the ellipse `(x^2)/(50)+(y^2)/(20)=1`from which a pair of perpendicular tangents is drawn to the ellipse `(x^2)/(16)+(y^2)/9=1`is0 (b) 2(c) 1 (d)4 |
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Answer» For the ellipse `(x^(2))/(16)+(y^(2))/(9)=1` the equation of director circle is `x^(2)+y^(2)=25`. The director circle will cut the ellipse `(x^(2))/(50)+(y^(2))/(20)=1` at four points. Hence, Number of points =4 |
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| 2. |
Tangents are drawn from the points on the line x-y-5=0 ot `x^(2)+4y^(2)=4` . Prove that all the chords of contanct pass through a fixed point |
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Answer» Variable point on the line x-y-=5 can be taken as (t,t-5) t `in R`. Chord of contact of the ellipse `x^(2)+4y^(2)=4 ` w.r.t. this point is `tx+4(t-5)y-4=0` or `(-20y-4)+t(x+4y)` This is the equation of family of straigth lines, Each member of this family passes through the point of intersection of straigth lines `-20y-4=0 and x+y0` which is `((1)/(5),-(4)/(5))` |
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| 3. |
If radius of the director circle of the ellipse `((3x+4y-2)^(2))/(100)+((4x-3y+5)^(2))/(625) =1` isA. 6B. `sqrt(34)`C. `sqrt(29)`D. `sqrt(26)` |
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Answer» Correct Answer - C `((3x+4y-2)^(2))/(100) +((4x-3y+5)^(2))/(625) =1` `rArr(((3x+4y-2)/(5))^(2))/(4)+(((4x-3y+5)/(5))^(2))/(25)=1` `rArr a^(2) =4` and `b^(2) = 25` Radius of director circle `= sqrt(a^(2)+b^(2)) = sqrt(29)` |
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| 4. |
PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R are `2alpha, 2beta, 2 gamma`, respectively then `tan beta gamma` is equal toA. `cot alpha`B. `cot^(2)alpha`C. `2 cot alpha`D. None of these |
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Answer» Correct Answer - B `(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1` `P(a cos 2alpha, b sin 2 alpha), Q (a cos 2 beta, b sin 2 beta)` `R(a cos 2 gamma,b sin 2 gamma)` Equation of chord PQ is `(x)/(a) cos (alpha + beta) + (y)/(b) sin (alpha + beta) = cos (alpha - beta)` PQ passes through the focus (ae,0) `:. e = (cos(alpha-beta))/(cos(alpha+beta))` `:. (cos(alpha-beta))/(cos(alpha+beta)) =-(cos(alpha-gamma))/(cos(alpha+gamma))` Apply componendo and dividendo, we get `(cos(alpha+beta)+cos(alpha-beta))/(cos(alpha+beta)-cos(alpha-beta))=(cos(alpha+gamma)-cos(alpha-gamma))/(cos(alpha+gamma)+cos(alpha-gamma))` `(2 cos alpha cos beta)/(2 sin alpha sin beta) = (2 sin alpha sin gamma)/(2 cos alpha cos gamma)` `tan beta tan gamma = cot^(2) alpha` |
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| 5. |
If the chords of contact of tangents from twopoinst `(x_1, y_1)`and `(x_2, y_2)`to theellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`are atright angles, then find the value of `(x_1x_2)/(y_1y_2)dot` |
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Answer» Correct Answer - `-a^(4)//b^(4)` The equations of the chords of contact of tangents drawn from `(x_(1),y_(1)) and (x_(2),y_(2))` to the ellipse `(x^(2))/(b^(2))+(y^(2))/(b^(2))=1` are `(x x_(1))/(a^(2))+(yy_(1))/(b^(2))=1" "(1)` `(x x_(2))/(a^(2))+(yy_(2))/(b^(2))=1" "(2)` It is given that (1) and (2) are at right angles. Therefore, `(-b^(2))/(a^(2))(x_(1))/(y_(1))xx(-b^(2))/(a^(2))(y_(2))/(y_(2))=-1` or `(x_(1)y_(1))/(y_(1)y_(2))=-(a^(2))/(b^(4))` |
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| 6. |
Tangents are drawn from any point on the circle `x^(2)+y^(2) = 41` to the ellipse `(x^(2))/(25)+(y^(2))/(16) =1` then the angle between the two tangents isA. `(pi)/(4)`B. `(pi)/(3)`C. `(pi)/(6)`D. `(pi)/(2)` |
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Answer» Correct Answer - D Given circle is director circle, so angle between tangents is `(pi)/(2)`. |
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| 7. |
Prove that the chords of constant of perpendicular tangents to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` touch another fixed ellipse `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/((a^(2)+b^(2)))` |
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Answer» We know, that the locus of the point of intersection of perpendicular tangents to the given ellipse is `x^(2)+y^(2)=a^(2)+b^(2)`. Any point on this circle can be taken as `P-=(sqrt(a^(2)+b^(2))cos theta,sqrt(a^(2)+b^(2))sin theta)` The equation of the chord of conatact of tangents from P is `(x)/(a^(2))sqrt(a^(2)+b^(2)) cos theta+(y)/(b^(2))sqrt(a^(2)+b^(2))sin theta=1` Let this line be a tangent be a tangne to the fixed ellipse `(x^(2))/(A^(2))+(y^(2))/(B^(2))=1` `:. (x)/(A) cos theta +(y)/(B) sin theta=1` Where `A=(a^(2))/(sqrt(a^(2)+b^(2))),B=(b^(2))/(sqrt(a^(2)+b^(2)))` Hence, the ellipe is `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/((a^(2)+b^(2)))` |
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| 8. |
Prove that the chords of contact of pairs of perpendicular tangents to the ellipse `x^2/a^2+y^2/b^2=1` touch another fixed ellipse.A. `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =(1)/((2a^(2)+b^(2)))`B. `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =(2)/((a^(2)-b^(2)))`C. `(x^(2))/(a^(4))+(y^(2))/(b^(4)) =(1)/((a^(2)+b^(2)))`D. `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =(2)/((3a^(2)-b^(2)))` |
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Answer» Correct Answer - C We known that locus of the point of intersection of perpendicular tangents to the given ellipse is `x^(2) + y^(2) = a^(2) + b^(2)`. Any point on this circle can be taken as `P -= (sqrt(a^(2)+b^(2)) cos theta, sqrt(a^(2)+b^(2)) sin theta)` The equation of the chord of contact of tangents from P is `(x)/(a^(2)) sqrt(a^(2)+b^(2)) cos theta +(y)/(b^(2)) sqrt(a^(2)+b^(2)) sin theta =1`. Let this line be a tangent to the fixed ellipse `(x^(2))/(A^(2)) +(y^(2))/(B^(2)) =1`. `rArr (x)/(A) cos theta +(y)/(B) sin theta =1`, Where `A = (a^(2))/(sqrt(a^(2)+b^(2))), B = (b^(2))/(sqrt(a^(2)+b^(2)))` `(x^(2))/(a^(4)) + (y^(2))/(b^(4)) = (1)/((a^(2)+b^(2)))` |
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| 9. |
Find the equation of the ellipse (referred to itsaxes as the axes of `xa n dy`, respectively) whose foci are `(+-2,0)`andeccentricity is `1/2` |
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Answer» Let the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Given`e=(1)/(2)` Also, foci of ellipse are `(+-ae,0)-=(+-2,0)` `rArr ae=2 rArr a=4` Now, `b^(2)=a^(2)(1-e^(2)) rArr b^(2) =12` Therefore, equation of ellipse is `(x^(2))/(16)+(y^(2))/(12)=1` |
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| 10. |
Find the equation of the ellipse, with major axis along the x-axis and passing through the points `(4, 3)`and `(-1, 4)`. |
| Answer» Correct Answer - ` 7x^(2)+15y^(2)= 247` | |
| 11. |
AB and CD are two equal and parallel chords of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1`. Tangents to the ellipse at A and B intersect at P and tangents at C and D at Q. The line PQA. passes through the originB. is bisected at the originC. cannot pass through the originD. is not bisected at the origin |
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Answer» Correct Answer - A::B Let P and Q be the points `(alpha, beta)` and `(alpha_(1),beta_(1))` `rArr` Equations of Ab and CD are `(x)/(a) alpha + (y)/(b) beta =1` and `(x)/(a) alpha_(1) +(y)/(b) beta_(1) =1` (Chord of contact) These lines are parallel `rArr (alpha)/(alpha_(1)) = (beta)/(beta_(1)) =k` Also `(alpha^(2))/(alpha^(2)) +(beta^(2))/(b^(2)) = (alpha_(1)^(2))/(a^(2)) + (beta_(1)^(2))/(b^(2))` `rArr (alpha)/(alpha_(1)) = (beta)/(beta_(1)) =-1` `rArr PQ` passes through origin and is bisected at the origin. |
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| 12. |
Find the equation of the ellipse whose foci are at `(pm 1, 0) and e=1/2.` |
| Answer» Correct Answer - `x^(2)/4+y^(2)/3 = 1` | |
| 13. |
Find the equation of the ellipse whose vertices are at `(0, pm 4) ` and foci at ` (0, pm sqrt7).` |
| Answer» Correct Answer - `x^(2)/9 + y^(2) / 16 = 1` | |
| 14. |
The length of the major axis of an ellipse is 20 units and its foci are (±5√3, 0) Find the equation of the ellipse. |
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Answer» Foci of the equation is in the form (±c, 0) Equation of the ellipse will be of the form: x2/a2 + y2/b2 = 1 Whose vertices (±a, 0) and foci (±c, 0) Here, c = 5√3 Length of the major axis of an ellipse is 20 units (given) So, length of the semi-major axis = 20/2 = 10 units, which is the value of a. ⇒ a = 10 Find b: We know, c2 = a2 – b2 (5√3)2 = 100 – b2 or b2 = 100- 75 = 25 Equation (1)⇒ x2/100 + y2/25 = 1 Which is required equation. |
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| 15. |
The length of the major axis of an ellipse is 20 units and its foci are `(pm 5sqrt3, 0).` Find the equation of the ellipse. |
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Answer» Correct Answer - `x^(2)/100+y^(2)/25 = 1` Let the equation of the ellipse be `x^(2)/a^(2) + y^(2)/b^(2) = 1.` Length of the major axis = ` 20 hArr 2a = 20 hArr a = 10.` Foci are `(pm c, 0) rArr c = 5 sqrt 3.` Now, ` c^(2) = (a^(2)-b^(2)) hArr b^(2) = (a^(2)-c^(2)) = (100-75) = 25.` `:. ` the required equation is `x^(2)/100+ y^(2)/25 = 1.` |
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| 16. |
Suppose `xa n dy`are real numbers and that `x^2+9y^2-4x+6y+4=0`. Then the maximum value of `((4x-9y))/2`is__________ |
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Answer» `x^(2)+9y^(2)-4x+6y+4=0` or `x^(2)-4y+9y^(2)+6y+4=0` or `(x-2)^(2)+(3y+1)^(2)=1` or `(x-2)^(2)+({y+(1//3)}^(2))/(1//9)=1` which is the equation of the ellipse having center at (2,-1//3). General point on the ellipse is `P(x,y)-=(2+a cos theta, -(1)/(3)+b sin theta)` `-=(2 + cos theta, -(1)/(3)+(1)/(3) sin theta)` `x=2+cos theta and y =-(1)/(3)+(1)/(3) sin theta` `:. 4x-9y=4(2+cos theta)-9(-(1)/(3)+(1)/(3)sin theta)` `or f(theta)=8+4cos theta+3-3 sin theta` `=11+4=16` `:. f(theta)_("max")=11+5=16` |
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| 17. |
`P and Q` are points on the ellipse `x^2/a^2+y^2/b^2 =1` whose center is `C`. The eccentric angles of P and Q differ by a right angle. If `/_PCQ` minimum, the eccentric angle of P can be (A) `pi/6` (B) `pi/4` (C) `pi/3` (D) `pi/12`A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(12)` |
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Answer» Correct Answer - B Since the eccentric angles of P and Q differ by a right angle, we can take P as `(a cos theta, b sin theta)` and Q as `(-a sin theta, b cos theta)` Slope of `CP = (b sin theta)/(a cos theta)` Slope of `CQ =-(b cos theta)/(a sin theta)` If Q is the angle between CP and CQ `A = tan^(-1) |((bsin theta)/(a cos theta)+(b cos theta)/(a sin theta))/(1-(b^(2)sin theta cos theta)/(a^(2)cos theta sin theta))| = tan^(-1) |(2ab)/(a^(2)-b^(2))-(1)/(sin 2 theta)|` Q is minimum if `sin 2 theta` is maximum. i.e., if `2 theta = (pi)/(2)` or `theta = (pi)/(4)` |
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| 18. |
For the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1` and `(x^(2))/(b^(2))+(y^(2))/(a^(2)) =1`A. The foci of each ellipse always lie within the other ellipseB. Their auxiliary circles are the nameC. Their director circles are the sameD. The ellipses encloses the same area |
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Answer» Correct Answer - B::C::D Without loss of generality assume `a gt b` Foci of `1^(st)` ellipse are `(+- ae, 0)` Putting this point in `(x^(2))/(b^(2)) + (y^(2))/(a^(2)) -1`, we get `(a^(2)e^(2))/(b^(2)) -1` The above quantity may be negative or positive, hence option (a) is not correct Auxiliary circle for `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) =1` is `x^(2) + y^(2) =a^(2)` for `(x^(2))/(b^(2)) + (y^(2))/(a^(2)) =1` is `x^(2) + y^(2) =a^(2)` Director circle for `(x^(2))/(b^(2)) +(y^(2))/(a^(2)) =1` is `x^(2)+y^(2) = a^(2) +b^(2)` `(x^(2))/(b^(2)) +(y^(2))/(a^(2)) =1` is Area of the ellipse `= pi ab` |
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| 19. |
Find the equation of the ellipse with foci at `(+-5,0)` and x=`36/5`as one of the directrices. |
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Answer» `(-ae,0)`and`(ae,0)` `ae=5+11` `a/e=36/5` `a^2=36` `a=pm6` `6*e=5` `e=5/6` `x^2/a^2+y^2/b^2=1` `x^2/a^2+y^2/(a^2(1-e^2))=1` `x^2/36+y^2/(36-25)=1` `x^2/36+y^2/11=1`. |
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| 20. |
Find the distance between the directrices ofthe ellipse `(x^2)/(36)+(y^2)/(20)=1.` |
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Answer» `x^2/36+y^2/20=1` `a^2=36,b^2=20` `a=pm6,b=pm2sqrt5` `e=sqrt(36-20)/6=2/3` `e=2/3` Distance between directrics=6/(2/3)=9. |
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| 21. |
`S_1a n dS_2`are the foci of an ellipse of major axis of length 10 units, and `P`is any point on the ellipse such that the perimeter of triangle `P S_1`is 15. Then the eccentricity of the ellipse is0.5 (b) 0.25 (c) 0.28 (d)0.75A. `0.5`B. `0.25`C. `0.28`D. `0.75` |
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Answer» `2a(1+e)=15` ltbr.or 1+e=(3)/(2)` or e=0.5 |
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| 22. |
P and Q are two points on the ellipse `(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1` whose eccentric angles are differ by `90^(@)`, thenA. Locus of point of intersection of tangents at P and Q is `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =2`B. Locus of mid-point `(P,Q)` is `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =(1)/(2)`C. Product of slopes of OP and OQ ehere O is the centre is `(-b^(2))/(a^(2))`D. Max. area of `DeltaOPQ` is `(1)/(2)ab` |
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Answer» Correct Answer - A::B::C::D `P =(a cos theta, b sin theta)` and `Q = (-a sin theta, b cos theta)` Tangent at `P, (x cos theta)/(a) + (y sin theta)/(b) =1` (1) `(-x sin theta)/(a) + (y cos theta)/(b) =1` (2) Elimination `theta`, by squaring and adding `(1)^(2) + (2)^(2) rArr (x^(2))/(a^(2)) + (y^(2))/(b^(2)) =2`, which is required locus. `:.` (a) is correct Now mid `(PQ) = ((a(cos theta-sin theta))/(2)(b(sin theta+cos theta))/(2)) =(x,y)` `cos theta - sin theta = (2x)/(a)` (3) `cos theta + sin theta = (2y)/(b)` (4) Squaring and adding (3) and (4), we get locus as `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) =(1)/(2)` `:.` (b) is correct Slope of `OP = (b sin theta)/(a cos theta) = m_(1)` Slope of `OQ = (-b cos theta)/(a sin theta) = m_(2)` `:. m_(1)m_(2) = (-b^(2))/(a^(2))` `:.` (c) is correct Now are of triangle `OPQ = (1)/(2) |{:(a cos theta, b sin theta),(-a sin theta,b cos theta):}|=(1)/(2)ab(cos^(2)theta+sin^(2)theta)=(1)/(2)ab` `:.` (d) is corrext |
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| 23. |
The tangent at any point on the ellipse `16x^(2)+25y^(2) = 400` meets the tangents at the ends of the major axis at `T_(1)` and `T_(2)`. The circle on `T_(1)T_(2)` as diameter passes throughA. `(3,0)`B. `(0,0)`C. `(0,3)`D. `(4,0)` |
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Answer» Correct Answer - A `(x^(2))/(5^(2)) +(y^(2))/(4^(2)) =1` Any tangent to the ellipse is `(x cos theta)/(5) + (y sin theta)/(4) =1` This meets `x = a =5` at `T_(1) {5,(4)/(sin theta) (1-cos theta)}` `= {5,4 tan.(theta)/(2)}` and meets `x =- a =- 5` at `T_(2) {-5,(4)/(sin theta) (1+cos theta)} = {-5,4 cot.(theta)/(2)}` The circle on `T_(1),T_(2)` as diameter is `(x-5) (x+5) + (y-4tan.(theta)/(2)) = 0` `x^(2) + y^(2) - 4y (tan.(theta)/(2)+cot.(theta)/(2)) -25 +16 =0` This is obviously satisfied by (3,0). |
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| 24. |
From a point P perpendicular tangents PQ and PR are drawn to ellipse `x^(2)+4y^(2) =4`, then locus of circumcentre of triangle PQR isA. `x^(2)+y^(2) =(16)/(5)(x^(2)+4y^(2))^(2)`B. `x^(2)+y^(2) =(5)/(16)(x^(2)+4y^(2))^(2)`C. `x^(2)+4y^(2)=(16)/(5)(x^(2)+y^(2))^(2)`D. `x^(2)+4y^(2)=(16)/(5)(x^(2)+y^(2))^(2)` |
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Answer» Correct Answer - B P lies on `x^(2) +y^(2) =5` (director circle of given ellipse) Let `P(sqrt(5) cos theta, sqrt(5) sin theta)` Since `DeltaQPR` is right angled at point P, circumcenter is mid-point of QR. QR is chord of contact w.r.t point P, `:.` Its equation is `(xsqrt(5)cos theta)/(4) +(y sqrt(5) sin theta)/(1) =1` (i) Also equation of chord QR which is bisected at point (h,k) is `(xh)/(h^(2) + 4k^(2)) +(4ky)/(h^(2)+4k^(2)) =1` (using `T = S_(1))` (ii) Comparing coefficients of equations (i) and (ii), `(sqrt(5)cos theta)/(4) = (h)/(h^(2) + 4k^(2))` and `sqrt(5) sin theta = (4k)/(h^(2)+4k^(2))` `rArr x^(2)+y^(2) = (5)/(16) (x^(2) + 4y^(2))^(2)` |
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| 25. |
Find the equation of the ellipse whose axes are oflength 6 and `2sqrt(6)`and theirequations are `x-3y+3=0`and `3x+y-1=0`, respectively. |
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Answer» Length of major axiz is 6 which is along the line x-3y+3=0 and lengthof axis is `2sqrt(6)` which is along the line 3x+y-1=0 Thus, equation of ellipse is `(((3x+y-1)/(sqrt(9+1)))/(3))^(2)+(((x-3y+3)/(sqrt(1+9)))/(sqrt(6)))^(2)=1` or `((3x+y-1))/(90)+((x-4y+3)^(2))/(60)=1` `rArr2(3x+y-1)^(2)+3(x-3y+3)^(2)=180` `rArr 21x^(2)-6xy+29y^(2)+6y-58y-151=0` |
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| 26. |
Let `Q = (3,sqrt(5)),R =(7,3sqrt(5))`. A point P in the XY-plane varies in such a way that perimeter of `DeltaPQR` is 16. Then the maximum area of `DeltaPQR` isA. 6B. 12C. 18D. 9 |
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Answer» Correct Answer - B P lies on the ellipse for which Q and R are foci such that `QR = 2ae =6` and `2a =` perimeter of triangle `-2ae = 16 - 6 = 10` `:. a = 5` and `e = 3//5`, also `b^(2) = a^(2) -a^(2) e^(2) = 25 -9 = 16` `b = 4` Maximum area of triangle `=(1)/(2) (2ae)(b) = 12` sq. units |
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| 27. |
The eccentricity of an ellipse whose centre is atthe origin is `1/2dot`if one of its directrices is `x=-4,`then the equation of the normal to it at `(1,3/2)`is:`4x+2y=7`(2) `x+2y=4`(3) `2y-x=2`(4) `4x-2y=1` |
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Answer» Eccentricity of the given ellipse`, e = 1/2` Equation of the directrix, `x = -4`. `:. +- a/e = -4` `=> +-a/(1/2) = -4` `=> a = +-2` `=>a^2 = 4` Now, we know, `e^2 = 1- b^2/a^2` `=>1/4 = 1-b^2/4` `=>b^2/4 = 3/4` `=>b^2 = 3` Now, equation of normal to an ellipse at point `(x_1,y_1)` is given by, `(a^2x)/(x_1) -(b^2y)/(y_1) = a^2-b^2` So, the equation of normal to the given ellipse at point `(1,3/2)` will be, `4x/1-3y/(3/2) = 4-3` `=>4x-2y = 1.` |
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| 28. |
If the normal at any point P on ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1` meets the auxiliary circle at Q and R such that `/_QOR = 90^(@)` where O is centre of ellipse, thenA. `a^(4) +2b^(3) ge 3a^(2)b^(2)`B. `a^(4) +2b^(4) ge 5a^(2)b^(2)+2a^(3)b`C. `a^(4)+2b^(4) ge 3a^(2)b^(2)+ab`D. None of these |
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Answer» Correct Answer - B Normal at `P(a cos theta, b sin theta)` is `ax sec theta - by cosec theta = a^(2) -b^(2)` Homogenising with auxilliary circle `x^(2) + y^(2) = a^(2)` `x^(2) + y^(2) = (a)^(2) ((ax sec theta - by cosec theta)^(2))/((a^(2)-b^(2))^(2))` `:.` For `/_QOR = 90^(@)` Coefficient of `x^(2)+` Coefficient of `y^(2) =0` `1- (a^(4)sec^(4)theta)/((a^(2)-b^(2))^(2)) + 1-(a^(2)b^(2)cosec^(2)theta)/((a^(2)-b^(2))^(2)) =0` `a^(4) - 5a^(2)b^(2) + 2b^(4) = a^(4) tan^(2) theta + a^(2)b^(2) cot^(2) theta` `:. AM ge GM` `a^(4) - 5a^(2)b^(2)+2b^(4) ge 2a^(3)b` |
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| 29. |
Find the equation of the common tangent in the first quadrant of the circle `x^2+y^2=16` and the ellipse `x^2/25+y^2/4=1`.Also find the length of the intercept of the tangent between the coordinates axes. |
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Answer» Correct Answer - `y=-(2x)/(sqrt(3))+4sqrt((7)/(3)),(14sqrt(3))/(3)` Let the common tangent to ` x^(2) + y^(2) = 16 and ( x^(2))/(25)+(y^(2))/(4)=1` be `y= mx +4 sqrt( 1+m^(2))` `and y= mx + sqrt(25 m^(2)+4)` since ,Eqs ((i) and (ii) are some tangent . `therefore 4sqrt(1+m^(2))=sqrt(25m^(2)+4)` `implies 16(1+m^(2))=25m^(2)+4` `implies 9m^(2)=12` `implies m=+- 2//sqrt(3)` since , tangent is in Ist quadrant . `therefore mlt0` `implies m=-2//sqrt(3)` so , the equation of the common tangent is `y=-(2x)/(sqrt(3))+4sqrt((7)/(3))` whcih meets coordinatae axes at `A (2 sqrt(7) ,0) and (0,4,sqrt((7)/(3)))` `therefore AB= sqrt((2sqrt(7)-0)^(2)+(0-4sqrt((7)/(3)))^(2))` `= sqrt(28+(11)/(3))=sqrt((196)/(3))=(14)/(sqrt(3))xx(sqrt(3))/(sqrt(3))=(14sqrt(3))/(3)` |
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| 30. |
Statement1: An equation of a common tangent to the parabola `y^2=16sqrt(3)x`and the ellipse `2x^2+""y^2=""4""i s""y""=""2x""+""2sqrt(3)`.Statement 2:If the line `y""=""m x""+(4sqrt(3))/m ,(m!=0)`is a common tangent to theparabola `y^2=""16sqrt(3)x`and the ellipse `2x^2+""y^2=""4`, then m satisfies `m^4+""2m^2=""24`.(1)Statement 1 isfalse, statement 2 is true(2)Statement 1 istrue, statement 2 is true; statement 2 is a correct explanation for statement1(3)Statement 1 istrue, statement 2 is true; statement 2 is not a correct explanation forstatement 1(4)Statement 1 istrue, statement 2 is falseA. Statement 1 is false 2 is trueB. Statement 1 is true, statement 2 true , statemens 2 is a correct explanation for statement 1C. Statement 1 is true, statement 2 is true: statement 2 is not a correct explanation for statement 1D. Statement 1 is true, statement 2 is false. |
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Answer» Equation of tangent to the ellipse `(x^(2))/(2)+(y^(2))/(4)=1` is `y=m x(4sqrt(3))/(m) " "(1)` Equation of tangent to the parabola `y^(2)=16sqrt(3)x` is `y=m x+(4sqrt(3))/(m)" "(2)` On comparing (1) and (2) `rArr48=m^(2)(2m^(2)+4)rArrm^(4)+2m^(2)-24=0` `rArr(m^(2)+6)(m^(2)-4)=0rArr m^(2)=4rArr m=+-2` So, equation of commn tangents are `y=+-2x+-2sqrt(3)` Statement 1 is true Statement 2 is obviously true and correct explanation of statement 1 |
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| 31. |
The area of the parallelogram formed by the tangents at the points whose eccentric angles are `theta, theta +(pi)/(2), theta +pi, theta +(3pi)/(2)` on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1` isA. abB. 4abC. 3abD. 2ab |
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Answer» Correct Answer - B Put `theta = 0^(@)`, we get rectangle formed by tangents at the extremities of major and minor axis. |
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| 32. |
The eccentricity of the ellipse `(x -3)^2 + (y-4)^2=y^2/9`A. `(sqrt(3))/(2)`B. `(1)/(3)`C. `(1)/(3sqrt(2))`D. `(1)/(sqrt(3))` |
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Answer» Correct Answer - B `9(x-3)^(2) +9(y-4)^(2) =y^(2)` `rArr 9(x-3)^(2) + 8y^(2) - 72y +144 =0` `rArr 9 (x-3)^(2) + 8(y^(2) -9y) +144 =0` `rArr 9 (x-3)^(2) + 8 [(y-(9)/(2))^(2)-(81)/(4)] +144 =0` `rArr 9(x-3)^(2) + 8 (y-(9)/(2))^(2) = 162- 144 = 18` `rArr(9(x-3)^(2))/(18) +(8(y-(9)/(2)))/(18) =1` `rArr ((x-3)^(2))/(2) + ((y-(9)/(2)))/(9//4) =1` `rArr e^(2) = 1 -(2xx 4)/(9) = (1)/(9) rArr e = (1)/(3)` |
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| 33. |
If normal at any poin P on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (agtbgt0)` meets the major and minor axes at Q and R, respectively, so that 3PQ=2PR, then find the eccentricity of ellipse |
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Answer» Correct Answer - `(1)/(sqrt(3))` Normal at point P meets the major axis minor axes at Q and R, respectively. `:. (PQ)/(PR)=(b^(2))/(a^(2))` Given `(PQ)/(PR)=(2)/(3)` `:. (b^(2))/(a^(2))=(2)/(3)` `:. E^(2)=1-(b^(2))/(a^(2))=1-(2)/(3)=(1)/(3)` `rArre=(1)/(sqrt(3))` |
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| 34. |
If the normal at any point `P`on theellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`meets theaxes at `Ga n dg,`respectively, then find the raio `P G: Pgdot` |
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Answer» Let `P( a cos theta, b sin theta)` be a point on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Then the quation of the normal at P is `ax sec theta-"by cosec" theta=a^(2)-b^(2)` It meets the axe at `G((a^(2)-b^(2))/(a)cos, theta,0)and g (0,-(a^(2)-b^(2))/(b) sin theta)` `:. PG^(2)=(acos theta-(a^(2)-b^(2))/(a)costheta)^(2)+b^(2)sin^(2)theta=(b^(2))/(a^(2))(b^(2) cos^(2)theta +a^(2)sin^(2)theta)` and `Pg^(2)=(a^(2))/(b^(2))(b^(2)cos^(2)theta +a^(2)sin^(2)theta)` `:. PG ,Pg=b^(2):a^(2)` |
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| 35. |
If the normal at any point P on the ellipse cuts the major and minor axes in G and g respectively and C be the centre of the ellipse, thenA. `a^(2)(CG)^(2)+b^(2)(Cg)^(2)=(a^(2)-b^(2))^(2)`B. `a^(2)(CG)^(2)-b^(2)(Cg)^(2) =(a^(2)-b^(2))^(2)`C. `a^(2)(CG)^(2)-b^(2)(Cg)^(2) =(a^(2)+b^(2))^(2)`D. None of these |
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Answer» Correct Answer - A At a point `(x_(1),y_(1))` on ellipse, normal will be `((x-x_(1))a^(2))/(x_(1)) =((y-y_(1))b^(2))/(y_(1))` At `G, y = 0 rArr x = CG = (x_(1)(a^(2)-b^(2)))/(a^(2))` At `g, x = 0 rArr y = Cg = (y_(1)(b^(2)-a^(2)))/(b^(2))` Now, `(x_(1)^(2))/(a^(2)) +(y_(1)^(2))/(b^(2)) = 1 rArr a^(2) (CG)^(2) + b^(2) (Cg)^(2) = (a^(2)-b^(2))^(2)` |
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| 36. |
Find the point on the ellipse `16 x^2+11 y^2=256`where thecommon tangent to ti and the circle `x^2+y^2-2x=15`toch. |
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Answer» The given ellipse is `(x^(2))/(16)+(y^(2))/((256//11))=1` Tangent to it at point `P( 4 cos theta, (16 sqrt(11)) sin theta) "is" (x)/(4) cos theta+(y)/((16//sqrt(11)))sin theta=1` It also touches the circle `(x-1)^(2)+y^(2)=16`. Therefore, `(|(1)/(4)cos theta-1|)/(sqrt((cos^(2)theta)/(16)+(11 sin^(2) theta)/256))=4` `or (|cos theta-4|)/(sqrt(16 cos^(2) theta+11 sin ^(2) theta))=1` `or cos^(2)theta-8costheta+16=11+5cos^(2)theta` or `4 cos^(2)theta+8cos theta-5=0` or `(2 cos theta-1)(2 cos theta+5)=0` or `cos theta =(1)/(2) or theta=+-(pi)/(3)` Therefore, point are `(2,+-sqrt(3)//sqrt(11))`. |
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| 37. |
Consider an ellipse `(x^2)/4+y^2=alpha(alpha`isparameter `>0)`and aparabola `y^2=8x`. If a common tangent to the ellipse and theparabola meets the coordinate axes at `Aa n dB`, respectively, then find the locus of themidpoint of `A Bdot` |
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Answer» The equation of tangent to `y^(2)=8x "at" (2t^(2),4t)` is `yt-x=2t^(2)=0" "(1)` The equation of tangent to the ellipse `(x^(2))/(4alpha)+(y^(2))/(alpha)=1` or `(2 sqrt(alpha) cos, theta, sqrt(alpha sin theta))` ltbrlt is ` (xcos theta)/(2sqrt(alpha))+(ysin theta)/(alpha))=1" "(2)` Comparing (1)and (2), we get `(sqrt(alpha))/(cos theta)=-t^(2),(sqrtalpha)/(sin theta)=2t" "(3)` Let the midpoint of AB be (h,k).Them, `h=(sqrt(alpha))/(cos theta),k=(sqrt(alpha))/(2 sin theta)` `:. h=-t^(2),k=t or k^(2)=-h` or `y^(2)=-x" "` [From (3)] |
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| 38. |
If the normal at any point P of the ellipse `(x^(2))/(16)+(y^(2))/(9) =1` meets the coordinate axes at M and N respectively, then `|PM|: |PN|` equalsA. `4:3`B. `16:9`C. `9:16`D. `3:4` |
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Answer» Correct Answer - C The equation of the normal at `P(theta)` on the ellipse is `4x sec theta - 3y cosec theta =7` This meets the coordinate axes at `M ((7)/(4) cos theta, 0), N (0,-(7)/(3) sin theta)` `:. PM^(2) = (4-(7)/(4))^(2) cos^(2) theta + 9 sin^(2) theta` `= (9)/(16) (9 cos^(2) theta + 16sin^(2) theta)` `PN^(2) = 16 cos^(2) theta + (3+(7)/(3))^(2) sin theta` `= (16)/(9) (9 cos^(2) theta + 16 sin^(2) theta)` `:. PM^(2): PN^(2) = 9^(2): 16^(2)` `rArr |PM| : |PN| = 9: 16` |
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| 39. |
If A and B are foci of ellipse `(x-2y+3)^(2)+(8x +4y +4)^(2) =20` andP is any point on it, then `PA +PB =`A. 2B. 4C. `sqrt(2)`D. `2sqrt(2)` |
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Answer» Correct Answer - B `(x-2y + 3)^(2) + (8x +4y +4)^(2) = 20` or `(((x-2y+3)/(sqrt(5))))/(4)+(((2x+y+1)/(sqrt(5)))^(2))/((1)/(4)) =1` `rArr PA + PB = 2a = 4` |
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| 40. |
If the tangent to the ellipse x2 + 4y2 = 16 at the point P (θ) is a normal to the circle x2 + y2 – 8x – 4y = 0 then θ equals(a) π/2(b) π/4(c) 0(d) -π/4 |
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Answer» Correct option (a) , (c) Explanation: Given ellipse is x2/16 + y2/4 = 1 Equation of tangent at P (θ) is P(acosθ ,bsinθ) i.e. P (4cosθ , 2sinθ) is 4x cosθ/16 + 2ysin θ/4 ....(1) xcosθ + 2ysinθ = 4 .......(1) (1) is a normal to the circle x2 + y2 – 8x – 4y = The equation (1) passes through the centre (4,2) of the circle. 4cosθ + 4sinθ = 4 cosθ + sinθ = 1 squaring 1 + sin2 = 1 sin2θ = 0 2θ = 0 or π Hence, θ = 0 or π |
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| 41. |
lf a quadrilateral is formed by four tangents to the ellipse `x^2/9+y^2/4=1` then the area of the square is equal to |
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Answer» Correct Answer - 26. sq. units When the quadrilateral is square, all the four vartices of square would line on the director circle of the ellipse. So, square is inscribed in the cirlce `x^(2)+y^(2)=a^(2)+b^(2)`. `:.` Area of square `=2(a^(2)+b^(2))=2(9+4)=26` sq. unit |
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| 42. |
If x2/f(4a) + y2/f(a2 - 5) = 1 represents an ellipse with major axis as y – axis and f is a decreasing function then(a) a∈(-∞, 1)(b) a∈(5,∞)(c) a∈(1,4)(d) a∈(-1, 5) |
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Answer» Correct option (d) a∈(-1, 5) Explanation: x2/f(4a) + y2/f(a2 - 5) = 1 represents an ellipse f(a2 – 5) > f(4a) ( ∴ major axis is y - axis ∴ b > a) Since f is decreasing function ∴ a2–5 < 4a (f(x1) > f (x2) then x1 < x2 for decreasing function) a2 – 4a – 5 < 0 a2 – 5a + a – 5< 0 (a – 5) (a + 1) < 0 a∈(-1, 5) |
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| 43. |
Find the foci of the ellipse `25(x+1)^2+9(y+2)^2=225.` |
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Answer» Correct Answer - `(-1,2)(-1,-6)` The given equation can be wirtten as `((x+1)^(2))/(9)+((y+2)^(2))/(25)=1` which represents an ellipse whose center is `(-1,-2`) and semi-major and the minor are 5 and 3, respectively. The eccentricity of the ellipse is given by `9=251-e^(2)or e=(4)/(5)` Shifting the origin at (-1,-2), the given reduces to `(x^(2))?(9)+(y^(2))/(25)=1" "(1)` where `x=X-1,y=Y-2" "(2)` The coordinates of the foci (1) are `(X=0,Y=+-be)`, where `b=5, e=4//5, i.e., the foci of (1) (X=0,Y+-4)`. Therefore, the cooridnates of the foci of the given ellipse are (-1,-2) and (-6,-6) |
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| 44. |
Find the equation of the normal to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`at the positive end of the latus rectum. |
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Answer» Correct Answer - `x-ey-e^(3)a=0` The equation of the normal at `(x_(1),y_(2))`, to the given ellipse is `(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)` Here, `x_(1)` =ad and `y_(1)=b^(2)//a` So, the equatio of the normal at the positive end of the latus rectum is `(a^(2)x)/(ae)-(b^(2)y)/(b^(2)//a)=a^(2)e^(2)" "[ :. b^(2)=a^(2)(1-e^(2))]` or `(ax)/(e)-ay=^(2)e^(2)or x-ey-e^(3)a=0` |
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| 45. |
The value of `a`for the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1,(a > b),`if the extremities of the latus rectum of the ellipse having positiveordinates lie on the parabola `x^2=2(y-2)`is ___ |
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Answer» `(+-ae,b^(2)//a)` are extermities of the latus rectum having positive ordinates. Then. `a^(2)e^(2)=-2((b^(2))/(a)-2) " "(1)` But `b^(2)=a^(2)(1-e^(2))" "(2)` Therefore, from (1) and (2), we get `a^(2)e^(2)-2ae^(2)-4=0` or `ae^(2)(a-2)+2(a-2)=0` `:. (ae^(2)+2)(a-2)=0` Hence, a=2 |
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| 46. |
The line `y=m x-((a^2-b^2)m)/(sqrt(a^2+b^2m^2))`is normal to the ellise `(x^2)/(a^2)+(y^2)/(b^2)=1`for all values of `m`belonging to`(0,1)`(b) `(0,oo)`(c) `R`(d) none of theseA. (0,1)B. `(0,oo)`C. RD. none of these |
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Answer» The eqquation of the normal to the given ellipse at the point `P(a cos theta, b sin theta) "is" ax sec theta- "by cosec" theta=a^(2)-b^(2)`. Then, `y=((a)/(b) tan theta)x-((a^(2)-b^(2)))/(b) sin theta" "(1)` Let `(a)/(b) tan theta=m` so that `y=((a)/(b) tan theta)x-((a^(2)-b^(2)))/(b) sin theta" "(1)` Hence, the equation of the normal equation (1) becomes `y=mx-((a^(2)-b^(2)))/(sqrt(a^(2)+b^(2)m^(2)))` ltbrlt `:. m in R, "as" m=(a)/(b) tan theta in R` |
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| 47. |
The number of distinct normal lines that can be drawn to the ellipse `(x^2)/(169)+(y^2)/(25)=1`from the point `P(0,6)`isone (b) two(c) three (d) fourA. oneB. twoC. threeD. four |
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Answer» `(x^(2))/(169)+ (y^(2))/(25=1` The equation of normal at the point `(13 cos theta, 5 sin theta)` is `(13x)/(cos theta)-(5y)/(sin theta)=144` It passes through (0,6). Thereofore, `0-30 =144 sin theta` `or sin theta=-(5)/(24)` or `theta=2pi-sin^(-1).((5)/(24)) and pi+sin^(-1).(5)/(24)` Also, the y-axis is one of the normals. |
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| 48. |
From the focus `(-5,0)` of the ellipse `(x^(2))/(45)+(y^(2))/(20) =1`, a ray of light is sent which makes angle `cos^(-1)((-1)/(sqrt(5)))` with the positive direction of X-axis upon reacting the ellipse the ray is reflected from it. Slope of the reflected ray isA. `-(3)/(2)`B. `-(7)/(3)`C. `-(5)/(4)`D. `-(2)/(11)` |
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Answer» Correct Answer - D Let `theta = cos^(-1) ((-1)/(sqrt(5))) rArr cos theta = (-1)/(sqrt(5)) rArr tan theta =-2` Foci are `(+- 5,0)` Equation of line through `(-5,0)` with slope `-2` is `y =- 2(x+5)` or `y =- 2x -10` This line meets the ellipse above X-axis at `(-6,2)` Reflected ray passes through the other focus `(5,0)` `:.` Slope `= (2-0)/(-6-5) =- (2)/(11)` |
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| 49. |
Find the (i) lengths of major and minor axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. `25x^(2)+4y^(2)=100` |
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Answer» Correct Answer - (i) 10 units, 4 units (ii) `A(0, -5) and B(0, 5) ` (iii) ` F_(1) (0, -sqrt(21)) and F_(2) (0, sqrt(21))" "` (iv) ` e= sqrt(21)/5 " "` (v) ` 1 3/5` units |
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| 50. |
Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse `9x^2+4y^2=36`. |
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Answer» Correct Answer - (i) 12 units, 4 units (ii) `A(0, -6) and B(0, 6)` (iii) `F_(1) (0, -4sqrt2) and F_(2) (0, 4sqrt2)" "` (iv) ` e= (2sqrt2)/3 " " ` (v) ` 1-1/3 ` units |
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