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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.
| 51. |
If `P(x_1,y_1),Q(x_2,y_2),R(x_3,y_3) and S(x_4,y_4)` are four concyclic points on the rectangular hyperbola ) and `xy = c^2` , then coordinates of the orthocentre ofthe triangle `PQR` is |
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Answer» We know that the orthocentre of the triangle formed by points `P(t_(1)), Q(t_(2)) and R(t_(3))` on hyperbola lies on the same hyperbola and has coordinates `((-c)/(t_(1)t_(2)t_(3)),-ct_(1)t_(2)t_(3))`. Also, if points `P(t_(1)), Q(t_(2)),R(t_(3)) and S(t_(4))` are concyclic then `t_(1)t_(2)t_(3)t_(4)=1.` Therefore, orhtocentre of triangle PQR is `(-ct_(4),(-c)/(t_(4)))-=(-x_(4),-y_(4)).` |
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| 52. |
If the circle `x^2+y^2=a^2` intersects the hyperbola `xy=c^2` in four points `P(x_1,y_1)`,`Q(x_2,y_2)`,`R(x_3,y_3)`,`S(x_4,y_4)`, then which of the following need not hold.(a) `x_1+x_2+x_3+x_4=0`(b) `x_1 x_2 x_3 x_4=y_1 y_2 y_3 y_4=c^4`(c) `y_1+y_2+y_3+y_4=0`(d) `x_1+y_2+x_3+y_4=0`A. `x_(1)+x_(2)+x_(3) +x_(4)=0`B. `y_(1)+y_(2)+y_(3) +y_(4)=0`C. `x_(1)x_(2)x_(3)x_(4)=c^(4)`D. `y_(1)y_(2)y_(3)y_(4)=c^(4)` |
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Answer» It is given that, `x^(2)+y^(2)=a^(2) " …(i)" ` and ` xy=c^(2) " …(ii)" ` We obtain ` x^(2)+c^(4)//x^(2)=a^(2)` `rArr x^(4)-a^(2)x^(2)+c^(4)=0 " …(iii)" ` Now `x_(1),x_(2),x_(3),x_(4)` will be root of Eq. (iii). Therefore, `Sigma x_(1)=x_(1)+x_(2)+x_(3)+x_(4)=0` and product of the roots `x_(1)x_(2)x_(3)x_(4)=c^(4)` Similarly, `y_(1)+y_(2)+y_(3)+y_(4)=0` and ` y_(1)y_(2)y_(3)y_(4)=c^(4)` Hence, all options are correct. |
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| 53. |
if `y=mx+7sqrt(3)` is normal to `(x^(2))/(18)-(y^(2))/(24)=1` then the value of m can beA. `(3)/(sqrt(5)`B. `(sqrt(15))/(2)`C. `(2)/(sqrt(5))`D. `(sqrt(5))/(2)` |
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Answer» Given equation of hyperbola, is `(x^(2))/(24)-(y^(2))/(18)=1 " …(i)" ` Since, the equation of the normals of slope m to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, are given by `y=mxpm(m(a^(2)+b^(2)))/(sqrt(a^(2)-b^(2)m^(2)))` `therefore ` Equation of normals of slope m, to the hyperbola (i), are `y=mx pm (m(24+18))/(sqrt(24-m^(2)(18)))" ...(ii)" ` `because " Line " y=mx+7sqrt(3)` is normal to hyperbola (i) ` therefore ` On comparing with Eq. (ii), we get ` pm (m(42))/(sqrt(24-18m^(2)))=7sqrt(3)` `rArr pm(6m)/(sqrt(24-18m^(2)))=sqrt(3)` `rArr (36m^(2))/(24-18m^(2))=sqrt(3) ` [squaring both sides] `rArr 12m^(2)=24-18m^(2)` `rArr 30m^(2)=24` `rArr 5m^(2)=4 rArr m=pm (2)/(sqrt(5))` |
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| 54. |
Let C be a curve which is locus of the point of the intersection of lines x = 2 + m and my = 4 – m. A circle (x – 2)2 + (y + 1)2 = 25 intersects the curve C at four points P, Q, R and S. If O is the centre of curve ‘C’ than OP2 + OQ2 + OR2 + OS2 is(a) 25(b) 50(c) 25/2(d) 100 |
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Answer» Correct option (d) 100 Explanation: x – 2 = m y + 1 = 4/m (x–2)(y+1) = 4 XY = 4, where X = x – 2 and Y = y + 1 and (x – 2)2 + (y + 1)2 = 25 X2 + Y2 = 25 Curve C and circle both are concentric. ∴ OP2 + OQ2 + OR2 + OS2 = 4r2 = 4 (25) = 100 |
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| 55. |
The length of the transverse axis of the rectangular hyperbola `x y=18`is6 (b) 12 (c) 18(d) 9A. `6`B. `12`C. `18`D. `9` |
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Answer» We know that the lengths of transverse and conjugate axes of the rectangular hyperbola `xy=(a^(2))/(2)` are each equal to `2a`. Here, `(a^(2))/(2)=18impliesa^(2)=36impliesa=6` Hence, length of the transverse axis `=12` |
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| 56. |
The equation of the transverse axis of the hyperbola(x – 3)2 + (y + 1)2 = (4x + 3y)2 is (a) 3x – 4y = 0(b) 4x + 3y = 0(c) 3x – 4y = 13(d) 4x + 3y = 9 |
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Answer» Correct option (c) 3x – 4y = 13 Explanation: (x – 3)2 + (y + 1)2 = (4x + 3y)2 (x – 3)2 + (y + 1)2 = 25 (4x + 3y/5)2 PS = 5 PM Directrix is 4x + 3y = 0 and focus is (3,–1) Equation of transverse axis is y + 1 = 3/4(x - 3) 3x - 4y = 13 |
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| 57. |
Find the product of the length of perpendiculars drawn from any pointon the hyperbola `x^2-2y^2-2=0`to its asymptotes.A. `1//2`B. `2//3`C. `3//2`D. `2` |
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Answer» We have, `(x^(2))/((sqrt(2))^(2))-(y^(2))/(1)=1`........`(i)` We know that the product of the lengths of perpendicular from any point on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` to its asymptotes is `(a^(2)b^(2))/(a^(2)+b^(2))` Here, `a^(2)=2` and `b^(2)=1`. So, required product `=(2)/(2+1)=(2)/(3)` |
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| 58. |
If angle between asymptotes of hyperbola x2/a2 - y2/b2 = 1 is 120° and product of perpendiculars drown from foci upon its any tangent is 9, then locus of point of intersection of perpendicular tangents of the hyperbola can be(a) x2 + y2 = 18(b) x2 + y2 = 6(c) x2 + y2 = 9(d) x2 + y2 = 3 |
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Answer» Correct option (a) x2 + y2 = 18 Explanation: 2tan–1 b/a = 60° b/a = 1/√3 b2 = 9 ∴ a2 = 27 Required locus is director circle i.e x2 + y2 = 27 – 9 x2 + y2 = 18 If b/a = tan 60° = √3 a2 = 3 Then equation of director circle is x2 + y2 = 3 - 9 = -6 which is not possible. |
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| 59. |
Find the equation of hyperbola whose foci are (±5, 0) and the conjugate axis is of the length 8. Also, find its eccentricity. |
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Answer» Foci of the equation is in the form (±c, 0) Equation of the hyperbola will be of the form: x2/a2 – y2/b2 = 1 Whose foci at (±c, 0) and conjugate axis is 2b On comparing, we get 2b = 8 ⇒ b = 4 And c = 5 Again, We know, c2 = a2 + b2 25 = 16 + a2 or a2 = 9 Equation (1)⇒ x2/9 – y2/16 = 1 Which is required equation. Again, Eccentricity (e) = c/a = 5/3 |
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| 60. |
Find the eccentricity of the hyperbola `x^(2) - y^(2) = 1`. If S, `S_(1)` are the foci and P any point on this hyperbola, prove that, `CP^(2) = SP*S_(1)P` (C is the origin.) |
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Answer» Correct Answer - `e=sqrt(2)` |
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| 61. |
Tangents are drawn to the hyperbola `x^2/9-y^2/4=1` parallet to the sraight line `2x-y=1.` The points of contact of the tangents on the hyperbola are (A) `(2/(2sqrt2),1/sqrt2)` (B) `(-9/(2sqrt2),1/sqrt2)` (C) `(3sqrt3,-2sqrt2)` (D) `(-3sqrt3,2sqrt2)`A. `((9)/(2sqrt2),(1)/(sqrt2))`B. `(-(9)/(2sqrt2),-(1)/(sqrt2))`C. `(3sqrt3,-2sqrt2)`D. `(3sqrt3,-2sqrt2)` |
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Answer» Correct Answer - A::B Slope of tangent = 2 The tangent are `y=2xpmsqrt(9xx4-4)" "("using y"=mx pmsqrt(a^(2)m^(2)-b^(2)))` `"i.e., "2x-y= pm 4sqrt2` `"or "(x)/(2sqrt2)-(y)/(4sqrt2)=1 and (x)/(2sqrt2)-(y)/(4sqrt2)=-1` Comparing it with `(x x_(1))/(9)-(yy_(1))/(4)=1` (Eqn. of tangent to hypebola at point `(x_(1),y_(1))` on it we get the point of contact as `(9//2sqrt2,1//sqrt2) and (-9//2sqrt2, -1//sqrt2)`. Alternatye Mathod: The equation of tangent at `P(theta)` is `((sec theta)/(3))x-((tan theta)/(2))y=1` `therefore" Slope"=(2 sec theta)/(3 tan theta)=2` `"or "sin theta=(1)/(3)` `therefore" "sec theta= pm(3)/(2sqrt2)` and corresponding by tan `theta= pm (1)/(2sqrt2)` Therefore, the points are `(9//2sqrt2,1//sqrt2)` and `(-9//2sqrt2, -1//sqrt2)`. |
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| 62. |
Find the eccetricity of hyperbola through `(4,2)` whose centre is `(0.0)` length of transverse axis is 4 and transverse axis along x-axis. (a) `2` (b) `sqrt3` (c) `(sqrt3)/(2)` (d) `(2)/(sqrt3)`A. 2B. `(2)/(sqrt(3))`C. `(3)/(2)`D. `sqrt(3)` |
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Answer» Equation of hyperbola is given by `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` ` because ` Length of transverse axis = 2a = 4 ` therefore a=2` Thus, `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is the equation of hyperbola ` because ` It passes through (4, 2). `therefore (16)/(4)-(4)/(b^(2))=1 rArr 4-(4)/(b^(2))=1 rArr b^(2)=(4)/(3) rArr b=(2)/(sqrt(3))` Now, eccentricity, `e=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+((4)/(3))/(4))=sqrt(1+(1)/(3))=(2)/(sqrt(3))` |
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| 63. |
Let `A` and `B` be two fixed points and `P`, another point in the plane, moves in such a way that `k_(1)PA+k_(2)PB=k_(3)`, where `k_(1)`, `k_(2)` and `k_(3)` are real constants. The locus of `P` is Which one of the above is not true ?A. a circle if `k_(1)=0` and `k_(2)`, `k_(3) gt 0`B. a circle if `k_(1) gt 0` and `k_(2) lt 0`, `k_(3) = 0`C. an ellipse if `k_(1)=k_(2) gt =0` and `k_(3) gt 0`D. a hyperbola if `k_(2)=-1` and `k_(1)`, `k_(3) gt 0` |
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Answer» If `k_(1)=0`, then `k_(1)PA+k_(2)PB=k_(3)` `impliesPB=(k_(3))/(k_(2)) gt 0` ,brgt `implies P` describes a circle with `B` as centre and radius `=(k_(3))/(k_(2))` If `k_(3)=0`, then `k_(1)PA+k_(2)PB=0` `implies(PA)/(PB)=(k_(2))/(-k_(1))= k gt 0` `impliesP` describes a circle with `P_(1)P_(2)` as its diameter, `P_(1)` and `P_(2)` being the points which divide `AB`. Internally and externally in the ratio `k : 1` If `k_(1)=k_(2) gt 0` and `k_(3) gt 0`, then `PA+PB=(k_(3))/(k_(1))=k gt 0` `impliesP` describes an ellipse with `A` and `B` as its foci. |
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| 64. |
The equation of a rectangular hyperbola whose asymptotes are `x=3`and `y=5`and passing through (7,8) is`x y-3y+5x+3=0``x y+3y+4x+3=0``x y-3y+5x-3=0``x y-3y+5x+3=0`A. `xy-3y+5x+3=0`B. `xy+3y+4x+3=0`C. `xy-3y+5x-3=0`D. `xy-3y-5x+3=0` |
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Answer» Correct Answer - D The equation of the rectangular hyperbola is `(x-3)(y-5)+lambda=0` It passes through (7, 8). Hence, `4xx3+lambda or lambda=-12` `therefore" "Xy-5x-3y+15-12=0` `"or "xy-3y-5x+3=0` |
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| 65. |
The equation of the line passing through the centre of a rectangular hyperbola is `x-y-1=0`. If one of its asymptotoes is `3x-4y-6=0`, the equation of the other asymptote isA. `4x+3y+17=0`B. `4x-3y+8=0`C. `3x-2y+15=0`D. none of these |
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Answer» The point of intersection of the line `x-y-1=0`, which passes through the centre of the hyperbola, and the asymptote `3x-4y-6=0` is the centre of the hyperbola. So, its coordinates are `(-2,-3)`. Since asymptotes of a rectangular hyperbola are always at right angle. So, required asymptotes is perpendicular to the given asymptote and passes through the centre `(-2,-3)` of the hypebola and hence its equation is `y+3=-(4)/(3)(x+2)` or, `4x+3y+17=0` |
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| 66. |
The director circle of a hyperbola is `x^(2) + y^(2) - 4y =0`. One end of the major axis is (2,0) then a focus isA. `(sqrt(3),2-sqrt(3))`B. `(-sqrt(3),2+sqrt(3))`C. `(sqrt(6),2-sqrt(6))`D. `(-sqrt(6),2+sqrt(6))` |
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Answer» Correct Answer - C::D Radius of director circle `sqrt(a^(2)-b^(2)) =2` Axis of hyperbola is line joining the center `C(0,2)` and `A(2,0)` (end of major axis) `:. a = CA = 2 sqrt(2)` `:. (2sqrt(2))^(2)-b^(2)=4` `:. b^(2) =4` `:. e = (sqrt(3))/(sqrt(2))` Center of the hyperbola is center of the director circle (0,2) Focus lies on this line at distance ae from center `:.` If foci are (x,y) then `(x-0)/(-(1)/(sqrt(2))) = (y-2)/((1)/(sqrt(2))) = +- 2 sqrt(3)` `(x,y) = (sqrt(6),2-sqrt(6))` or `(-sqrt(6),2+sqrt(6))` |
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| 67. |
Consider a hyperbola: `((x-7)^(2))/(a) -((y+3)^(2))/(b^(2)) =1`. The line `3x - 2y - 25 =0`, which is not a tangent, intersect the hyperbola at `H ((11)/(3),-7)` only. A variable point `P(alpha +7, alpha^(2)-4) AA alpha in R` exists in the plane of the given hyperbola. The eccentricity of the hyperbola isA. `sqrt((7)/(5))`B. `sqrt(2)`C. `(sqrt(13))/(2)`D. `(3)/(2)` |
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Answer» Correct Answer - C The given line must be parallel to asymptotes `rArr` Slope of asymptotes are `(3)/(2)` and `-(3)/(2)` `rArr (b)/(a) =(3)/(2) rArr e = (sqrt(13))/(2)` |
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| 68. |
The points on the ellipse `(x^(2))/(2)+(y^(2))/(10)=1` from which perpendicular tangents can be drawn to the hyperbola `(x^(2))/(5)-(y^(2))/(1) =1` is/areA. `(sqrt((3)/(2)),sqrt((5)/(2)))`B. `(sqrt((3)/(2)),-sqrt((5)/(2)))`C. `(-sqrt((3)/(2)),sqrt((5)/(2)))`D. `(sqrt((5)/(2)),sqrt((3)/(2)))` |
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Answer» Correct Answer - A::B::C Required points will lie on the intersection of ellipse `(x^(2))/(2)+(y^(2))/(10)=1` with director circle of hyperbola `(x^(2))/(5)-(y^(2))/(1) =1` i.e. on `x^(2) + y^(2) =4` `rArr (sqrt(2)cos theta)^(2) + (sqrt(10)sin theta)^(2) =4` Solving, we get `sin theta = +- (1)/(2), cos theta = +-(sqrt(3))/(2)` `:.` Points are `(+-sqrt((3)/(2)),+-sqrt((5)/(2)))` |
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| 69. |
If the foci of `(x^(2))/(16)+(y^(2))/(4)=1` and `(x^(2))/(a^(2))-(y^(2))/(3)=1` coincide, the value of a isA. 3B. 2C. `(1)/(sqrt(3))`D. `sqrt(3)` |
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Answer» Correct Answer - A Foci of `(x^(2))/(16)+(y^(2))/(4) =1` are `(+- sqrt(12),0)` Foci of `(x^(2))/(a^(2)) -(y^(2))/(3) =1` are `(+- sqrt(a^(2)+3), 0)` Given `a^(2)+3 =12 rArr a^(2) = 9 rArr a = 3` |
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| 70. |
(x-1)(y-2)=5 and `(x-1)^2+(y+2)^2=r^2` intersect at four points A, B, C, D and if centroid of `triangle ABC` lies on line `y = 3x-4` , then locus of D is |
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Answer» Given hyperbola is `(x-1)(y-2)=5" (1)"` and circle is `(x-1)^(2)+(y+2)^(2)=r^(2)" (2)"` These curves intersect at four points A, B, C and D. We know curves intersect at four points is midpoint of centres of curves. `(sum_(i=1)^4x_(i))/(4)=(1+1)/(2)=1 and (sum_(i-1)^(4)y_(1))/(4)=(2-2)/(2)=0,` where `(x_(i),y_(i),i=1,2,3,4` are points A, B, C and D. `therefore" "sum_(i=1)^(4)x_(1)=4 and sum_(i=1)^(4)y_(i)=0` `therefore" "x_(1)+x_(2)+x_(3)=4-x_(4) and y_(1)+y_(2)+y_(3)=-y_(4)` So, centroid of triangle ABC is `((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))-=((4-x_(4))/(3),(-y_(4))/(3))` Now, given that centroid lies on the line `y = 3x-4`. `therefore" "(-y_(y))/(3)=3((4-x_(4))/(3))-4` `therefore" "y_(4)=3x_(4)` Therefore, y=3x, which is locus of point D. |
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| 71. |
Find the equations of the tangents to the hyperbola `x^2=9y^2=9`that are drawn from (3, 2). |
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Answer» The equation of the hyperbola is `(x^(2))/(9)-(y^(2))/(1)=1` The equation of the tangent haivng slope m is `y=mx pm sqrt(9m^(2)-1)` It passes through (3, 2). Therefore, `2=3m pm sqrt(9m^(2)-1)` `"or "4+9m^(2)-12m=9m^(2)-1` `"i.e., "m=(5)/(12)or m=oo` Hence, the equations of the tangents are `y-3=(5)/(12)(x-2) and x=3` |
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| 72. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola.Horizontal Hyperbola:For general form of Hyperbola:x2/a2 – y2/b2 = 1 ……..(1)Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2)x2/9 – y2/16 = 1 |
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Answer» Given equation is of the form x2/a2 – y2/b2 = 1 On comparing given equation with (1), we get a = 3 and b = 4 Then, c2 = a2 + b2 c2 = 9 + 16 = 25 or c = 5 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 6 units Length of Conjugate axis = 2b = 8 units (ii) Coordinates of the vertices: (±a, 0) = (±3, 0) (iii) Coordinates of the foci: (±c, 0) = (±5, 0) (iv) Eccentricity: e = c/a = 5/3 (v) Length of the latus rectum: 2b2/a = 2(16)/5 = 32/5 units |
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| 73. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) x2/25 – y2/4 = 1 |
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Answer» Given equation is of the form x2/a2 – y2/b2 = 1 On comparing given equation with (1), we get a = 5 and b = 2 Then, c2 = a2 + b2 c2 = 25 + 4 = 29 or c = √29 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 10 units Length of Conjugate axis = 2b = 4 units (ii) Coordinates of the vertices: (±a, 0) = (±5, 0) (iii) Coordinates of the foci: (±c, 0) = (±√29, 0) (iv) Eccentricity: e = c/a = √29/5 (v) Length of the latus rectum: 2b2/a = 8/5 units |
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| 74. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) x2 – y2 = 1 |
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Answer» Given equation is of the form x2/a2 – y2/b2 = 1 On comparing given equation with (1), we get a = 1 and b = 1 Then, c2 = a2 + b2 c2 = 1 + 1 = 2 or c = √2 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 2 units Length of Conjugate axis = 2b = 2 units (ii) Coordinates of the vertices: (±a, 0) = (±1, 0) (iii) Coordinates of the foci: (±c, 0) = (±√2, 0) (iv) Eccentricity: e = c/a = √2 (v) Length of the latus rectum: 2b2/a = 2 units |
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| 75. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) 3x2 – 2y2 = 6 |
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Answer» Divide each side by 6 x2/2 – y2/3 = 1 Given equation is of the form x2/a2 – y2/b2 = 1 On comparing given equation with (1), we get a = √2 and b = √3 Then, c2 = a2 + b2 c2 = 2 + 3 = 5 or c = √5 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 2√2 units Length of Conjugate axis = 2b = 2√3 units (ii) Coordinates of the vertices: (±a, 0) = (±√2, 0) (iii) Coordinates of the foci: (±c, 0) = (±√5, 0) (iv) Eccentricity: e = c/a = √(5/2) (v) Length of the latus rectum: 2b2/a = 3√2 units |
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| 76. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) y2/9 – x2/27 = 1 |
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Answer» Given equation is of the form y2/a2 – x2/b2 = 1 On comparing given equation with (2), we get a = 3 and b = √27 = 3√3 Then, c2 = a2 + b2 c2 = 9 + 27 = 36 or c = 6 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 6 units Length of Conjugate axis = 2b = 6√3 units (ii) Coordinates of the vertices: (0, ±a) = (0, ±3) (iii) Coordinates of the foci: (0, ±c) = (0, ±6) (iv) Eccentricity: e = c/a = 2 (v) Length of the latus rectum: 2b2/a = 18 units |
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| 77. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) y2/16 – x2/49 = 1 |
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Answer» Given equation is of the form y2/a2 – x2/b2 = 1 On comparing given equation with (2), we get a = 4 and b = 7 Then, c2 = a2 + b2 c2 = 16 + 49 = 65 or c = √65 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 8 units Length of Conjugate axis = 2b = 14 units (ii) Coordinates of the vertices: (0, ±a) = (0, ±4) (iii) Coordinates of the foci: (0, ±c) = (0, ±√65) (iv) Eccentricity: e = c/a = √65/4 (v) Length of the latus rectum: 2b2/a = 49/2 units |
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| 78. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) 24x2 – 25y2 = 600 |
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Answer» Divide both sides by 600 x2/25 – y2/24 = 1 Given equation is of the form x2/a2 – y2/b2 = 1 On comparing given equation with (1), we get a = 5 and b = 2√6 Then, c2 = a2 + b2 c2 = 25 + 24 = 49 or c = 7 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 10 units Length of Conjugate axis = 2b = 4√6 units (ii) Coordinates of the vertices: (±a, 0) = (±5, 0) (iii) Coordinates of the foci: (±c, 0) = (±7, 0) (iv) Eccentricity: e = c/a = 7/5 (v) Length of the latus rectum: 2b2/a = 48/5 units |
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| 79. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) 25x2 – 9y2 = 225 |
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Answer» Divide both sides by 225 x2/9 – y2/25 = 1 Given equation is of the form x2/a2 – y2/b2 = 1 On comparing given equation with (1), we get a = 3 and b = 5 Then, c2 = a2 + b2 c2 = 9 + 25 = 34 or c = √34 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 6 units Length of Conjugate axis = 2b = 10 units (ii) Coordinates of the vertices: (±a, 0) = (±3, 0) (iii) Coordinates of the foci: (±c, 0) = (±√34, 0) (iv) Eccentricity: e = c/a = √34/3 (v) Length of the latus rectum: 2b2/a = 50/3 units |
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| 80. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) 3y2 – x2 = 108 |
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Answer» Divide each side by 108 y2/36 – x2/108 = 1 Which is of the form y2/a2 – x2/b2 = 1 On comparing given equation with (2), we get a = 6 and b = 6√3 Then, c2 = a2 + b2 c2 = 36 + 108 = 144 or c = 12 Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 12 units Length of Conjugate axis = 2b = 12√3 units (ii) Coordinates of the vertices: (0, ±a) = (0, ±6) (iii) Coordinates of the foci: (0, ±c) = (0, ±12) (iv) Eccentricity: e = c/a = 2 (v) Length of the latus rectum: 2b2/a = 36 units |
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| 81. |
Find the (i) lengths of axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of the hyperbola. Horizontal Hyperbola:For general form of Hyperbola: x2/a2 – y2/b2 = 1 ……..(1) Vertical Hyperbola:For general form of Hyperbola:y2/a2 – x2/b2 = 1 ……..(2) 5y2 – 9x2 = 36 |
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Answer» Divide each side by 36, we get y2/(36/5) – x2/4 = 1 Which is of the form y2/a2 – x2/b2 = 1 On comparing given equation with (2), we get a = 6/√5 and b = 2 Then, c2 = a2 + b2 c2 = 36/5 + 4 = 56/5 or c = 2√(14/5) Now, (i) Lengths of the axes: Length of Transverse axis = 2a = 12/√5 units Length of Conjugate axis = 2b = 4 units (ii) Coordinates of the vertices: (0, ±a) = (0, ±6/√5) (iii) Coordinates of the foci: (0, ±c) = (0, ±2√(14/5)) (iv) Eccentricity: e = c/a = √14/3 (v) Length of the latus rectum: 2b2/a = 4√5/3 units |
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| 82. |
Find the equation of hyperbola whose vertices at (±6, 0) and foci at (±8, 0). |
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Answer» Vertices are of the form (±a, 0), hence it is a horizontal hyperbola. So, the equation is of the form: x2/a2 – y2/b2 = 1 …..(1) Given: vertices of hyperbola (±6, 0) Here a = 6 Again, Foci is of the form (±c, 0) Given: foci of hyperbola at (±8, 0) ⇒ c = 8 Find b: We know, c2 = a2 + b2 64 = 36 + b2 or b2 = 64 – 36 = 28 Equation (1)⇒ x2/36 – y2/28 = 1 Which is required equation. |
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| 83. |
Find the equation of the hyperbola with vertices at (0, ±5) and foci at (0, ±8). |
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Answer» Vertices are of the form (0, ±a), hence it is a vertical hyperbola. So, the equation is of the form: y2/a2 – x2/b2 = 1 …..(1) Given: vertices of hyperbola (0, ±5) Here a = 5 Again, Foci is of the form (0, ±c) Given: foci of hyperbola at (0, ±8) ⇒ c = 8 Find b: We know, c2 = a2 + b2 64 = 25 + b2 or b2 = 39 Equation (1)⇒ y2/25 – x2/39 = 1 Which is required equation. |
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| 84. |
Find the equation of hyperbola whose foci are (±√29, 0) and the transverse axis is of the length 10. |
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Answer» Major axis of the given hyperbola lie on x-axis. Equation of the hyperbola will be of the form: x2/a2 – y2/b2 = 1 Whose foci at (±c, 0) and transverse axis is 2a Given: transverse axis = 10 ⇒ 2a = 10 ⇒ a = 5 Given: foci (±√29, 0) (±c, 0) = (±√29, 0) ⇒ c = √29 We know, c2 = a2 + b2 29 = 25 + b2 or b2 = 4 Equation (1)⇒ x2/25 – y2/4 = 1 Which is required equation. |
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| 85. |
A common tangent to `9x^2-16y^2 = 144` and `x^2 + y^2 = 9`, is |
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Answer» Correct Answer - `y=pm3sqrt((2)/(7))xpm(15)/(sqrt7)` Given equation of hyperbola is `9x^(2)-16y^(2)=144` `"or "(x^(2))/(16)-(y^(2))/(9)=1` The equation of tangent to the hyperbola having slope m is brgt `y=mxpmsqrt(16m^(2)-9)` If it touches the circle, then the distance of the line from the centre of the circle is the radius of the circle. Hence, `(sqrt(16m^(2)-9))/(sqrt(m^(2)+1))=3` `"or "9m^(2)+9=16m^(2)-9` `"or "7m^(2)=18` `"or "m=pm3sqrt((2)/(7))` So, the equation of tangent is `y=pm3sqrt((2)/(7))xpm(15)/(sqrt7)` |
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| 86. |
`x^(2)-y^(2)=1` |
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Answer» Correct Answer - (i) 2 units , 2 units (ii) `A(-1,0) ,B(1,0)` (iii) `F_(1)(-sqrt(2),0),F_(2) (sqrt(2),0)` (iv) `e=sqrt(2)` (v) 2 units |
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| 87. |
`(x^(2))/(25)-(y^(2))/(4)=1` |
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Answer» Correct Answer - (i) 10 units , 4 units (ii) `A(-5,0),B(5,0)` (iii) `F_(1)(-sqrt(29),0),F_(2)(sqrt(29),0)` (iv) `e=(sqrt(29))/(5)` `(v) 1(1)/(3) ` units |
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| 88. |
`(x^(2))/(9)-(y^(2))/(16)=1` |
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Answer» Correct Answer - (i) 6 units , 8 units (ii) `A(-3,0),B(3,0) ` (iii) `F_(1) (-5,0),F_(2)(5,0)` `(iv) e=(5)/(3) ` ` (v) 10(2)/(3)` units |
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| 89. |
`25x^(2)-9y^(2)=225` |
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Answer» Correct Answer - (i) 6 units, 10 units (ii) `A(-3,0),B(3,0) ` (iii) `F_(1)(-7,0),F_(2)(sqrt(34),0)` (iv) `e=(sqrt(34))/(3)` (v) `16(2)/(3)` units |
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| 90. |
`5y^(2)-9x^(2)=36` |
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Answer» Correct Answer - `(i) (12)/(sqrt(5))"units,4units"` (ii) `A(0,-(6)/(sqrt(5))),B(0,(6)/(sqrt(5)))` `(iii) F_(1)(0,-2sqrt((14)/(5))),F_(2)(0,2sqrt((14)/(5)))` (iv) `e=(sqrt(14))/(3)` (v) `(4sqrt(5))/(3) ` units Given equation is `(y^(2))/((36//5))-(x^(2))/(4)=1.` `therefore a^(2)=(36)/(5),b^(2)=4 and c^(2)=(a^(2)+b^(2))=((36)/(5)+4)=(56)/(5).` Also `, e=( c)/(a)=((sqrt(56))/(sqrt(5))xx(sqrt(5))/(6))=(2sqrt(14))/(6)=(sqrt(14))/(3).` |
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| 91. |
Find the vertices of the hyperbola `9x^2=16 y^2-36 x+96 y-252=0`A. `(6,3)` and `(-6,3)`B. `(6,3)` and `(-2,3)`C. `(-6,3)` and `(-6,-3)`D. none of these |
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Answer» The equation of the hyperbola is `9(x^(2)-4x+4)-16(y^(2)-6y+9)=144` `implies((x-2)^(2))/(4^(2))-((y-3)^(2))/(3^(2))=1` The coordinates of the vertices are given by `x-2=+-4`, `y-3=0` `impliesx=6`, `y=3` and `x=-2`, `y=3` Hence, the coordinates of the vertices are `(6,3)` and `(-2,3)` . |
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| 92. |
`3x^(2)-2y^(2)=6` |
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Answer» Correct Answer - (i) `2 sqrt(2) " units," 2sqrt(3) "units "` (ii) `A(-sqrt(2),0),B (sqrt(2),0)` (iii) `F_(1)(-sqrt(5),0),F_(2)(sqrt(5),0)` `(iv) e=sqrt((5)/(2))` (v) `3sqrt(2)` units |
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| 93. |
The eccentricity of the hyperbola with latursrectum `12` and semi-conjugate axis is `2sqrt(3)`, isA. `2`B. `3`C. `sqrt(3)//2`D. `2sqrt(3)` |
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Answer» We have, `(2b^(2))/(a)=12` and, `b=2sqrt(3)impliesa=2` `:.e=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+(12)/(4))=2` |
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| 94. |
The equation of the tangent parallel to `y=x` drawn to `(x^(2))/(3)-(y^(2))/(2)=1`, isA. `x-y+1=0`B. `x-y+2=0`C. `x-y+3=0`D. `x-y-2=0` |
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Answer» Let `y=x+c` be the required tangent. Then, `c^(2)=3xx1^(2)-2impliesc=+-1` [ Using `c^(2)=a^(2)m^(2)-b^(2)`] Hence required tangents are `y=x+-1`. |
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| 95. |
The equation of the hyperbola with vertices `(3,0)` and `(-3,0)` and semi-latursrectum `4`, is given byA. `4x^(2)-3y^(2)+36=0`B. `4x^(2)-3y^(2)+12=0`C. `4x^(2)-3y^(2)-36=0`D. none of these |
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Answer» We have, `a=3` and `(b^(2))/(a)=4impliesb^(2)=12`. Hence, the equation of the hyperbola is `(x^(2))/(9)-(y^(2))/(12)=1implies4x^(2)-3y^(2)=36` |
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| 96. |
Find the equation of tangents to the curve `4x^2-9y^2=1`which are parallel to `4y=5x+7.`A. `24y-30x=17`B. `30y-24x=+-sqrt(161)`C. `24y-30x=+-sqrt(161)`D. none of these |
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Answer» Let `m` be the slope of the tangent to `4x^(2)-9y^(2)=1` Then, `m=("Slope of the line" 4y=5x+7)=5//4` We have, `(x^(2))/(1//4)-(y^(2))/(1//9)=1` or, `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, where `a^(2)=(1)/(4)`, `b^(2)=(1)/(9)` The equations of the tangents are `y=mx+-sqrt(a(2)m^(2)-b^(2))` `impliesy=(5)/(4)x+-sqrt((25)/(64)-(1)/(9))` `implies30x-24y+-sqrt(161)=0implies24y-30x=+-sqrt(161)` |
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| 97. |
Find the equation of tangents to the curve `4x^2-9y^2=1`which are parallel to `4y=5x+7.` |
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Answer» Let m be the slope of the tangent of `4x^(2)-9y^(2)=1.` Then `m=("Slope of the line "4y=5x+7)=(5)/(4)` We have `(x^(2))/(1//4)-(y^(2))/(1//9)=1` where `a^(2)=(1)/(4),b^(2)=(1)/(9)` The equations of the tangents are `y=mx pm sqrt(a^(2)m^(2)-b^(2))` `"or "y=(5)/(4)xpmsqrt((25)/(64)-(1)/(9))` `"or "30x-24y pm sqrt(161)=0` `"or "24y-30x=pmsqrt(161)` |
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| 98. |
The angle between pair of tangents to the curve `7x^(2)-12y^(2)=84` from the point `M(1,2)` isA. `2tan^(-1)"(1)/(2)`B. `2tan^(-1)2`C. `2(tan^(-1).(1)/(3)+tan^(-1).(1)/(2))`D. `2tan^(-1)3` |
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Answer» The equation of the director circles of the given byperbola is `x^(2)+y^(2)=12-7` or `x^(2)+y^(2)=5` Clearly, the point `M(1,2)` lies on the director circle `x^(2)+y^(2)=5`. So, the angle between the tangents to the hyperbola drawn from `M(1,2)` is right angle. Now, `2(tan^(-1).(1)/(3)+tan^(-1).(1)/2)=2tan^(1)(((1)/(3)+(1)/(2))/(1-(1)/(3)xx(1)/(2)))=2tan^(1)1=(pi)/(2)` Hence, option `(c )` is correct. |
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| 99. |
Find the equation of the hyperbola whose vertices are `(+-7,0)` and the eccentricity is `(4)/(3).` |
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Answer» Since the vertices of the given hyperbola are of the form `(+-a,0)`,it is a horizontal hyperbola . Let the requied equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1.` then , its vertices are `(+-a,0)`. But , the vertices are `(+-7,0)` `therefore a=7hArra^(2)=49.` Also ,`e=( c)/(a)hArrc=ae=(7xx(4)/(3))=(28)/(3).` Now , `C^(2)=(a^(2)+b^(2))hArrb^(2)=(c^(2)-a^(2))=[((28)/(3))^(2)-49]=(343)/(9).` Thus`,a^(2)=49 and b^(2)=(343)/(9).` ` therefore` the required equation is `(x^(2))/(49)-(y^(2))/((343//9))=1hArr(x^(2))/(49)-(9y^(2))/(343)=1hArr 7x^(2)-9y^(2)=343.` |
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| 100. |
Tangents are drawn to the hyperbola `x^2/9-y^2/4=1` parallet to the sraight line `2x-y=1.` The points of contact of the tangents on the hyperbola are (A) `(2/(2sqrt2),1/sqrt2)` (B) `(-9/(2sqrt2),1/sqrt2)` (C) `(3sqrt3,-2sqrt2)` (D) `(-3sqrt3,2sqrt2)`A. `(+-(9)/(2sqrt(2)),+-(1)/(sqrt(2)))`B. `(+-(1)/(sqrt(2)),+-(9)/(2sqrt(2)))`C. `(3sqrt(3),-2sqrt(2))`D. `(-3sqrt(3),2sqrt(2))` |
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Answer» The points of contact of tangents of slope `m` to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` are `(+-(a^(2)m)/(sqrt(a^(2)m^(2)-b^(2))),+-(b^(2))/(sqrt(a^(2)m^(2)-b^(2))))` Here , `m=2`, `a^(2)=9` and `b^(2)=4`. So, the points of contact are `(+-(18)/(sqrt(36-4)),+-(4)/(sqrt(36-4)))=(+-(9)/(2sqrt(2)),+-(1)/(sqrt(2)))` |
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