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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.
| 151. |
The coordinates of the vertices of the hyperbola`9x^(2)-16y^(2)=144` are-A. `(0,pm4)`B. `(pm4,0)`C. `(0,pm8)`D. `(pm8,0)` |
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Answer» Correct Answer - B |
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| 152. |
Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.`(y^2)/9-(x^2)/(27)=1` |
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Answer» Correct Answer - (i) 6 units , `6sqrt(3)` units (ii) `A(0,-4),B(0,3)` (iii) `F_(1)(0,-6),F_(2) (0,6)` (iv) `e=2` (v) 18 units |
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| 153. |
For the hyperbola `x^2/ cos^2 alpha - y^2 /sin^2 alpha = 1;(0 lt alphalt pi/4)`A. EccentricityB. Abscissa of fociC. DirectrixD. Vertex |
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Answer» Correct Answer - B `e^(2)=1+(b^(2))/(a^(2))=1+(sin^(2)alpha)/(cos^(2)alpha)=(1)/(cos^(2)alpha)` `a^(2)=cos^(2)alpha` `therefore" "a^(2)e^(2)=1` Hence, the foci are `(pmae, 0)-=(pm1,0)`, which are independent of `alpha.` |
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| 154. |
Find the equation of the hyperbola whose vertices are `(0,+-3)` and the eccentricity is `(4)/(3)` ,Also find the coordinates of its foci . |
| Answer» Correct Answer - `(y^(2))/(9)-(x^(2))/(7)=1,(0,+-4)` | |
| 155. |
Find the equation of the hyperbola whose eccentricity is `(5)/(3)` and whose vertices are `(0,+-6).` Also, find the coordinates of its foci . |
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Answer» Since the vertices of the given hyperbola are of the form `(-0,+-a)`, it is a vertical hyperbola . Let the required equation be `(y^(2))/(a^(2))-(x^(2))/(b^(2))=1.` Then , its vertices are `(0,+-a)` But , it is given that the vertices are `(0,+-6).` `therefore a=6.Also , e=(5)/(3).` Now , `(c )/(a)=eimpliesc=ae=(6xx(5)/(3))=10.` `And ,c^(2)=(a^(2)+b^(2))hArr b^(2)=(c^(2)-a^(2))={(10)^(2)-6^(2)}=(100-36)=64.` THus , `a^(2)=6^(2)=36 and b^(2)=64.` Hence , the required equation is `(y^(2))/(36)-(x^(2))/(64)=1.` Also , the coordinates of its foci are `(0,+-c),i.e., (0,+-10).` |
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| 156. |
The centre of a hyperbola is at (2,4), the coordinates of one of its vertices are (6,4) and eccentricity is `sqrt(5)`, find its equation. |
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Answer» Correct Answer - `4(x-2)^(2)-(y-4)^(2)=64` |
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| 157. |
The length of the transverse axis of the hyperbola `9y^(2) - 4x^(2) = 36` is -A. 2 unitB. 3 unitC. 4 unitD. 5 unit |
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Answer» Correct Answer - C |
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| 158. |
The length of latus rectum of the hyperabola `9y^(2) - 4x^(2) = 36` is -A. 8 unitB. 9 unitC. 10 unitD. 12 unit |
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Answer» Correct Answer - B |
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| 159. |
Which of the following pairs may represent the eccentricities of two conjugate hyperbolas, for `alpha in (0,pi//2)`?A. `sin theta, cos theta`B. `tan theta, cot theta`C. `sec theta, "cosec"theta`D. `1+sintheta,1+cos theta` |
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Answer» Correct Answer - C If `e_(1)` and `e_(2)` are the eccentricities of two conjugate hyperbolas, then `(1)/(e_(1)^(2))+(1)/(e_(2)^(2))=1` Therefore, `e_(1)=sec theta and e_(2)="cosec"theta.` |
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| 160. |
Find the equation of the hyperbola, whose axes are axes of coordinates and transverse axis is 2a and the vertex bisects the line segment joining the centre and focus. |
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Answer» Correct Answer - `3x^(2)-y^(2) = 3a^(2)` |
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| 161. |
Let `x^2/a^2+y^2/b^2=1 and x^2/A^2-y^2/B^2=1` be confocal `(a > A and a> b)` having the foci at`s_1 and S_2,` respectively. If P is their point of intersection, then `S_1 P and S_2 P` are the roots of quadratic equationA. `x^(2)+2ax+(a^(2)-A^(2))=0`B. `x^(2)+2ax+(a^(2)-A^(2))=0`C. `x^(2)-2Ax+(a^(2)+A^(2))=0`D. `x^(2)-2ax+(a^(2)-A^(2))=0` |
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Answer» Correct Answer - D `S_(1)+S_(2)P=2a and S_(1)P-S_(2)P=2A` `S_(1)+=a+A and S_(2)P=(a-A)` Required quadratic equation is `x^(2)-2ax+(a^(2)-A^(2))=0` |
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| 162. |
The coordinates of the foci of a hyperbola are (16,0) and (-8,0) and its eccentricity is 3, find the equation of the hyperbola and the length of its latus rectum. |
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Answer» Correct Answer - `8(x-4)^(2)-y^(2)=128 and 64`unit |
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| 163. |
The length of latus rectum of the hyperbola `9y^(2) - 4x^(2) = 36`A. `(pm2,0)`B. `(pm3,0)`C. `(0,pm2)`D. `(0,pm3)` |
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Answer» Correct Answer - C |
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| 164. |
If the distances of one focus of hyperbola from its directrices are 5 and 3, then its eccentricity isA. `sqrt2`B. 2C. 4D. 8 |
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Answer» Correct Answer - B We have `ae-(a)/(e)=3and ae+(a)/(e)=5` `therefore" "ae=4and (1)/(e)=1` `therefore" "a^(2)=4` `therefore" "a=2` `therefore" "e=2` |
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| 165. |
The equation of the transvers axis of the hyperbola `(x-3)^2+(y=1)^2+(4x+3y)^2`is`x+3y=0`(b) `4x+3y=9``3x-4y=13`(d) `4x+3y=0` |
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Answer» Correct Answer - `3x-4y=13` `(x-3)^(2)+(y+1)^(2)=(4x+3y)^(2)` `"or "sqrt((x-3)^(2)+(y+1)^(2))=5((|4x+3y|)/(sqrt(4^(2)+3^(2))))` Clearly, it is hyperbola with focus at (3, -1) and directrix `4x+3y=0`. Transverse axis is perpendicular to directrix and passes through focus. So, its equation is `y+1=(3)/(4)(x-3)` `"or "3x-4y=13` |
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| 166. |
Find the equation of the hyperbola, whose axes are axes of coordinates and eccentricity is `(sqrt(13))/(3)` and the sum of the squares of the lengths of axes is 52. |
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Answer» Correct Answer - `4x^(2)-9y^(2) = 36` |
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| 167. |
The distance between two directrices of a rectangular hyperbola is 10units. Find the distance between its foci. |
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Answer» Correct Answer - 20 Distance between the two directrices is `(2a)/(e)=10" units"` or a = 5e Now, distance between the foci is `2ac=10e^(2)=10(2)=20` `" "("As rectangular hyperbola, e" = sqrt2)` |
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| 168. |
The distance between the vertices of a hyperbola is half the distance between its foci and the length of its semi-conjugate axis is 6. Referred to its axes of coordinates, find the equation of the hyperbola. |
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Answer» Correct Answer - `3x^(2)-y^(2)=36` |
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| 169. |
Show that the equation 9`x^2-16 y^2-18 x+32 y-151=0`represents a hyperbola. Find the coordinates of the centre,lengths of the axes, eccentricity, latus-rectum,coordinates of foci and vertices, equations of the directricesof the hyperbola. |
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Answer» `9x^2-16y^2-18x+32y=151` `9(x^2-2x)-16(y^2-2y)=151` `9(x^2-9x+1)-16(y^2-2y+1)+9+16=151` `9(x-1)^2-16(y-1)^2=144` `(x-1)^2/16-(y-1)^2/9=1` `a^2=16,b^2=9` `Centre(1,1)` `b^2=a^2(e^2-1)` `b^2/a^2+1=e^2` `9/16+1=e^2` `e=5/4` `x-1=pma/e` `x=pma/e+1` `4*4/5+1` transverse`=2a=8` Conj=2b=6. |
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| 170. |
From any point to the hyperbola `^2/a^2-y^2/b^2=1`, tangents are drawn to thehyperbola `x^2/a^2-y^2/b^2=2` The area cut off bythe chord of contact on the regionbetween the asymptotes is equal toA. `a//2`B. `ab`C. `2ab`D. `4ab` |
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Answer» Correct Answer - D Let `P(x_(1),y_(1))` be a point on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` The chord of contact of tangents from P to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=2` is given by `(x x_(1))/(a^(2))-(yy_(1))/(b^(2))=2" "(1)` The equation of the asymptotes are `(x)/(a)-(y)/(b)=0` `and (x)/(a)+(y)/(b)=0` the points of intersection of (1) with the two asymptotes are given by `x_(1)=(2)/((x_(1)//a)-(y_(1)//b)),y_(1)=(2b)/((x_(1)//a)-(y_(1)//b))` `x_(2)=(2)/((x_(1)//a)-(y_(1)//b)),y_(2)=(-2b)/((x_(1)//a)-(y_(1)//b))` `"Area of the said triangle"=(1)/(2)|x_(1)y_(2)-x_(2)y_(1)|` `=(1)/(2)|-(4abxx2)/((x_(1)^(2)//a^(2))-(y_(1)^(2)//b^(2)))|=4ab` |
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| 171. |
The asymptotes of the hyperbola `(x^(2))/(a_(1)^(2))-(y^(2))/(b_(1)^(2))=1` and `(x^(2))/(a_(2)^(2))-(y^(2))/(b_(2)^(2))=1` are perpendicular to each other. Then,A. `a_(1)//a_(2)=b_(1)//b_(2)`B. `a_(1)a_(2)=b_(1)b_(2)`C. `a_(1)a_(2)+b_(1)b_(2)=0`D. `a_(1)-a_(2)=b_(1)-b_(2)` |
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Answer» Correct Answer - C The slopes of asymptotes are `m_(1)=(b_(1))/(a_(1)),m_(2)=(b_(2))/(a_(2))` According to the question, `m_(1)m_(2)=-1` `"or "a_(1)a_(2)+b_(1)b_(2)=0` |
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| 172. |
From any point to the hyperbola `^2/a^2-y^2/b^2=1`, tangents are drawn to thehyperbola `x^2/a^2-y^2/b^2=2` The area cut off bythe chord of contact on the regionbetween the asymptotes is equal toA. `(ab)/(2)`B. `ab`C. `2ab`D. `4ab` |
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Answer» Let `P(x_(1),y_(1))` be a point on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. Then, `(x_(1)^(2))/(a^(2))-(y_(1)^(2))/(b^(2))=1` The chord of contact of tangents from `P` to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=2` is `(x x_(1))/(a^(2))-(y y_(1))/(b^(2))=2`..........`(i)` The equations of the asymptotes are `(x)/(a)-(y)/(b)=0` and `(x)/(a)+(y)/(b)=0` The points of intersection of `(i)` with the two asymptotes are given by `x_(1)=(2a)/((x_(1))/(a)-(y_(1))/(b))`, `y_(1)=(2b)/((x_(1))/(a)-(y_(1))/(b))`, `x_(2)=(2a)/((x_(1))/(a)+(y_(1))/(b))`, `y_(2)=(2b)/((x_(1))/(a)+(y_(1))/(b))`, `:.` Area of the triangle `=(1)/(2)(x_(1)y_(2)-x_(2)y_(1))=(1)/(2)((4abxx2)/((x_(1)^(2))/(a^(2))-(y_(1)^(2))/(b^(2))))=4ab` |
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| 173. |
The eccentricity of the hyperbola whose length of the latus rectum isequal to 8 and the length of its conjugate axis is equal to half of thedistance between its foci, is :(1) `4/3`(2) `4/(sqrt(3))`(3) `2/(sqrt(3))`(4) `sqrt(3)`A. `3//4`B. `4//sqrt3`C. `2//sqrt3`D. none of these |
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Answer» Correct Answer - C We have `(2b^(2))/(a)=8` `and 2b=(1)/(2)(2ae)` `therefore" "(2)/(a)((ae)/(2))^(2)=8` `"or "ae^(2)=16" (1)"` Also, `2xx(b^(2))/(a)=8` `"or "b^(2)=4a` `"or "a^(2)(e^(2)-1)=4a` `"or "ae^(2)-a=4" (2)"` From (1) and (2), we have `16-(16)/(e^(2))=4` `"or "(16)/(e^(2))=12` `"or "e=(2)/(sqrt3)` |
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| 174. |
The eccentricity of the hyperbola whose length of the latus rectum isequal to 8 and the length of its conjugate axis is equal to half of thedistance between its foci, is :(1) `4/3`(2) `4/(sqrt(3))`(3) `2/(sqrt(3))`(4) `sqrt(3)`A. `(4)/(3)`B. `(4)/(sqrt(3))`C. `(2)/(sqrt(3))`D. `sqrt(3)` |
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Answer» We have, `(2b^(2))/(a)=8 and 2b=ae` ` rArr b^(2)=4a and 2b=ae` consider, `2b=ae` `rArr 4b^(2)=a^(2)e^(2)` `rArr 4a^(2)(e^(2)-1)=a^(2)e^(2)` `rArr 4e^(2)-4=e^(2) " "[ because a ne 0]` ` rArr 3e^(2)=4` `rArr e=(2)/(sqrt(3)) " "[because e gt 0]` |
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| 175. |
Write the length o the latus rectum of the hyperbola `16 x^2-9y^2=144.` |
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Answer» Correct Answer - `e=5//4, "Vertices"-=(0, pm4), "Foci"-=(0, pm5), L.R. = 9//2, "Directrix;y"=pm16//5` We have hyperbola `16x^(2)-9y^(2)=-144` `"or "(x^(2))/(9)-(y^(2))/(16)=-1` This equation is of the form `(x^(2))/(a^(2))-(y^(2))/(16)=-1`. Hence, x-axis is the conjugate axis and y-axis is the tranverse axis. Now, `a^(2)=9,b^(2)=16." So, "a=3,b=4.` Length of transverse axis = 2b = 8 Length of conjugate axis = 2a = 6 Eccentricity, `e=sqrt(1+(a^(2))/(b^(2)))=sqrt(1+(9)/(16))=(5)/(4)` Vertics are `(0 pmb) or (0, pm4)`. Foci are `(0, pm be) or (0, pm 5)` Length of latus rectum `=(2a^(2))/(b)=(2(3)^(2))/(4)=(9)/(2)` Equation of directrices are `y=pm(b)/(e)` `"or "y=pm(4)/((5//4))` `"or "y=pm(16)/(5)` |
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| 176. |
If the latus rectum of a hyperbola is equal to half of its transverse axis, then its eccentricity is -A. `sqrt(3)`B. `sqrt((3)/(2))`C. `(2)/(sqrt(2))`D. 2 |
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Answer» Correct Answer - B |
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| 177. |
If `H(x,y)=0` represents the equation of a hyperbola and `A(x,y)=0`, `C(x,y)=0` the joint equation of its asymptotes and the conjugate hyperbola respectively, then for any point `(alpha, beta)` in the plane `H(alpha,beta)`, `A(alpha,beta)` , and `C(alpha,beta)` are inA. `A.P.`B. `G.P.`C. `H.P.`D. none of these |
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Answer» Let `H(x,y)=(x^(2))/(a^(2))-(y^(2))/(b^(2))-1`. Then, `A(x,y)=(x^(2))/(a^(2))-(y^(2))/(b^(2))` and `C(x,y)=(x^(2))/(a^(2))-(y^(2))/(b^(2))+1` We observe that `2A(alpha,beta)=H(alpha,beta)+C(alpha,beta)` Hence, `H(alpha,beta)`, `A(alpha,beta)` and `C(alpha,beta)` are in `A.P`. |
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| 178. |
A tangent drawn to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =1`at `P((pi)/(6))` forms a triangle of area `3a^(2)` square units, with coordinate axes, then the squae of its eccentricity is equal toA. 15B. 16C. 17D. 18 |
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Answer» Correct Answer - C The point `P ((pi)/(6))` is `(asec.(pi)/(6),b tan.(pi)/(6))`, i.e., `P((2a)/(sqrt(3)),(b)/(sqrt(3)))` `:.` Equation of tangent at P is `(x)/((sqrt(3)a)/(2))-(y)/(sqrt(3)b) =1` `:.` Area of the triangle `=(1)/(2) xx (sqrt(3)a)/(2) xx sqrt(3) b = 3a^(2)` `:. (b)/(a) = 4` `:. e^(2) = 1 + (b^(2))/(a^(2)) = 17` |
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| 179. |
Normal is drawn at one of the extremities of the latus rectum of thehyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`which meets the axes at points `Aa n dB`. Then find the area of triangle `O A B(O`being the origin). |
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Answer» Correct Answer - `"Area"=(1)/(2)a^(2)e^(5)` Normal at point `P(x_(1),y_(1))` is `(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))=a^(2)+b^(2).` It meets the axes at `A(((a^(2)_b^(2))x_(1))/(a^(2)),0)and B(0,((a^(2)+b^(2))y_(1))/(b^(2)))` `"Area of "DeltaOAB=(1)/(2)[((a^(2)+b^(2))x_(1))/(a^(2))][((a^(2)+b^(2))y_(1))/(b^(2))]` `=(1)/(2)[((a^(2)+b^(2))x_(1)y_(1))/(a^(2)b^(2))]` Now, normal is drawn at the extremity of latus rectum. Hence, `(x_(1),y_(1))-=(ae,(b^(2))/(a))` `therefore" Area"=(1)/(2)[((a^(2)+b^(2))^(2)b^(2)e)/(a^(2)b^(2))]` `=(1)/(2)[(a^(4)(1+(b^(2))/(a^(2)))^(2)e)/(a^(2))]` `=(1)/(2)a^(2)e^(5)` |
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| 180. |
Find the equation of the hyperbola, referred to its axes as lines parallel to coordinate axes and having centre at (3,-2). Eccentricity `(sqrt(5))/(2)` and the length of latus rectum 2. |
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Answer» Correct Answer - `(x-3)^(2)-4(y+2)^(2)=16` |
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| 181. |
If `a x+b y=1`is tangent to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`, then `a^2-b^2`is equal to`1/(a^2e^2)`(b) `a^2e^2``b^2e^2`(d) none of theseA. `1//a^(2)e^(2)`B. `a^(2)e^(2)`C. `b^(2)e^(2)` none of theseD. none of these |
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Answer» Correct Answer - A Any tangent to hyperbola is `(x)/(a) sec theta-(y)/(b) tan theta=1" (1)"` The given tangent is `ax+by=1" (2)"` Comparing (1) and (2), we have `sec theta=a^(2)and tan theta=-b^(2)` Eliminating `theta`, we have `a^(4)-b^(4)=1` `"or "(a^(2)-b^(2))(a^(2)+b^(2))=1` Also, `a^(2)+b^(2)=a^(2)e^(2)` `"or "a^(2)-b^(2)=(1)/(a^(2)e^(2))` |
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| 182. |
If the tangent and normal to a rectangular hyperbola cut off intercepts `a_(1)` and `a_(2)` on one axis and `b_(1)` and `b_(2)` on the other, thenA. `a_(1)a_(2)+b_(1)b_(2)=0`B. `a_(1)a_(2)=-b_(1)b_(2)`C. `a_(1)b_(2)=a_(2)b_(1)`D. `a_(1)a_(2)=b_(1)b_(2)` |
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Answer» Let the equation of the rectangular hyperbola be `x^(2)-y^(2)=a^(2)` and, let `P(a sec theta, a tan theta)` be a point on it. The equations of the tangent and normal at `P` are `x sec theta-y tantheta=a`……..`(i)` and, `x cos theta+y cot theta=2a`..........`(ii)` It is given that the tangent at `P` cuts off intercepts `a_(1)` and `b_(1)` on the coordinate axes. Therefore, `a_(1)=a cos theta` and `b_(1)=-a cot theta` The normal at `P` cuts off intercepts `a_(2)` and `b_(2)` on the coordinate axes. Therefore, `a_(2)=2a sec theta` and `b_(2)=2a tan theta` Now, `a_(1)a_(2)+b_(1)b_(2)=a cos thetaxx2a sec thet+(-a cot theta)(2a tantheta)` `impliesa_(1)a_(2)+b_(1)b_(2)=2a^(2)-2a^(2)=0` |
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| 183. |
If `x=9` is the chord of contact of the hyperbola `x^2-y^2=9` then the equation of the corresponding pair of tangents is (A) `9x^2-8y^2+18x-9=0` (B) `9x^2-8y^2-18x+9=0` (C) `9x^2-8y^2-18x-9=0` (D) ` `9x^2-8y^2+18x+9=0`A. `9x^(2)-8y^(2)+18x-9=0`B. `9x^(2)-8y^(2)-18x=0`C. `9x^(2)-8y^(2)-9=0`D. `9x^(2)-8y^(2)+18x+9=0` |
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Answer» Correct Answer - B Let a pair of tangents be drawn from the point `(x_(1),y_(1))` to the hyperbola `x^(2)-y^(2)=9` Then the chord of contact will be `x x_(1)-yy_(1)=9" (1)"` But the given chord of contact is `x = 9" (2)"` As (1) and (2) represent the same line, these equations should be identical and, hence, `(x_(1))/(1)=-(y_(1))/(0)=(9)/(9)or x_(1)=1,y_(1)=0` Therefore, the equation of pair to tangents drawn from (1, 0) to `x^(2)-y^(2)=9` is `(x^(2)-y^(2)-9)(1^(2)-0^(2)-9)=(x*1-y*0-9)^(2)` `" "("Using SS"_(1)=T^(2))` `"or "(x^(2)-y^(2)-9)(-8)=(x-9)^(2)` `"or "-8x^(2)+8y^(2)+72=x^(2)-18x+81` `"or "9x^(2)-8y^(2)-18x+9=0` |
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| 184. |
The locus of a point, from where the tangents to the rectangularhyperbola `x^2-y^2=a^2`contain an angle of `45^0`, is`(x^2+y^2)^2+a^2(x^2-y^2)=4a^2``2(x^2+y^2)^2+4a^2(x^2-y^(2))=4a^2``(x^2+y^2)^2+4a^2(x^2-y^2)=4a^2``(x^2+y^2)+a^2(x^2-y^(2))=a^4`A. `(x^(2)+y^(2))^(2)+a^(2)(x^(2)-y^(2))=4a^(2)`B. `2(x^(2)+y^(2))^(2)+4a^(2)(x^(2)-y^(2))=4a^(2)`C. `(x^(2)+y^(2))^(2)+4a^(2)(x^(2)-y^(2))=4a^(4)`D. `(x^(2)+y^(2))^(2)+a^(2)(x^(2)-y^(2))=a^(4)` |
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Answer» Correct Answer - C Let `y=mx pm sqrt(m^(2)a^(2)-a^(2))` be two tangents that pass through (h,k). Then `(k-mh)^(2)=m^(2)a^(2)-a^(2)` `"or "m^(2)(h^(2)-a^(2))-2khm+k^(2)+a^(2)=0` `"or "m_(1)+m_(2)=(2kh)/(h^(2)-a^(2))` and `m_(1)m_(2)=(k^(2)+a^(2))/(h^(2)-a^(2))` Now, `tan45^(@)=(m_(1)-m_(2))/(1+m_(1)m_(2))` `"or "1=((m_(1)+m_(2))^(2)-4m_(1)m_(2))/((1+m_(1)m_(2))^(2))` `"or "(1+(k^(2)+a^(2))/(h^(2)-a^(2)))^(2)=((2kh)/(h^(2)-a^(2)))^(2)-4((k^(2)+a^(2))/(h^(2)-a^(2)))` `"or "(h^(2)+k^(2))^(2)=4h^(2)k^(2)-4(k^(2)+a^(2))(h^(2)-a^(2))` `"or "(x^(2)+y^(2))^(2)=4(a^(2)y^(2)-a^(2)x^(2)+a^(4))` `"or "(x^(2)+y^(2))^(2)+4a^(2)(x^(2)-y^(2))=4a^(4)` |
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| 185. |
If the chord `xcosalpha+ysinalpha=p`of the hyperbola `(x^2)/(16)-(y^2)/(18)=1`subtends a right angle at the center, and the diameter of the circle,concentric with the hyperbola, to which the given chord is a tangent is `d ,`then the value of `d/4`is__________ |
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Answer» Correct Answer - 24 The equation of hyperbola is `(x^(2))/(16)-(y^(2))/(18)=1` `"or "9x^(2)-8y^(2)-144=0` Homogenizing this equation using `(x cosalpha+y sin alpha)/(p)=1` We have `9x^(2)-8y^(2)-144((x cos alpha+y sin alpha)/(p))^(2)=0` Since these lines are perpendicular to each other, we have `9p^(2)-8p^(2)-144(cos^(2)alpha+sin^(2)alpha)=0` `p^(2)=144or p = pm12` `therefore" Radius of circle"=12` `therefore" Diameter of circle"=24` |
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| 186. |
If ` 5x +9 =0 ` is the directtrix of the hyperbola `16x^(2) -9y^(2)=144,` then its corresponding focus isA. `(-(5)/(3),0)`B. `(-5,0)`C. `((5)/(3),0)`D. `(5,0)` |
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Answer» Equation of given hyperbola is `16x^(2)-9y^(2)=144` `rArr (x^(2))/(9)-(y^(2))/(16)=1 " ...(i)" ` So, the eccentricity of Eq. (i) `e=sqrt(1+(16)/(9))=(5)/(3)` [` because ` the eccentricity (e) of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is `sqrt(1+(b//a)^(2))`] and given directrix is `5x+9=0 rArr x= -9//5` So, corresponding focus is `(-3((5)/(3)),0)=(-5,0)` |
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| 187. |
The equation of the chord of contact of tangents drawn from a point `(2,-1)` to the hyperbola `16x^(2)-9y^(2)=144`, isA. `9x+32y=144`B. `32x-9y=144`C. `32x+9y=144`D. none of these |
| Answer» The equation of the chord of contact of tangents drawn from `(2,-1)` to the hyperbola `16x^(2)-9y^(2)=144` is `32x+9y=144`. | |
| 188. |
The equation of the tangent to the hyperbola `16x^(2)-9y^(2)=144` at `(5,8//3)`, isA. `10x+3y=18`B. `10x-3y=18`C. `10x-3y=9`D. `10x+3y=9` |
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Answer» The equation fo the tangent at `(5,8//3)` is `16(5x)-9(8//3)=144implies10x-3y=18` |
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| 189. |
If a variable straight line `x cos alpha+y sin alpha=p` which is a chord of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 (b gt a)` subtends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius, isA. `(ab)/(sqrt(b-2a))`B. `(a)/(a-b)`C. `(ab)/(sqrt(b^(2)-a^(2)))`D. `(ab)/(sqrt(b+a))` |
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Answer» The combined equation of the straight lines joining, the centre of the hyperbola of the points of intersection of the line `x cosalpha+ysinalpha=p` and the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is `(x^(2))/(a^(2))-(y^(2))/(b^(2))=((xcosalpha+ysinalpha)/(p))^(2)` This is equation will represent a pair of perpendicular straight lines, if `(1)/(a^(2))-(cos^(2)alpha)/(p^(2))-(1)/(b^(2))-(sin^(2)alpha)/(p^(2))=0impliesp=(ab)/(sqrt(b^(2)-a^(2)))` Substituting the value of `p` in `x cosalpha+ysinalpha=p`, we get `x cosalpha+ysinalpha=(ab)/(sqrt(b^(2)-a^(2)))` Clearly, it touches the circle `x^(2)+y^(2)=(-(ab)/(sqrt(b^(2))-a^(2))^(2))` |
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| 190. |
The locus of a point whose chord of contact with respect to the circle `x^2+y^2=4`is a tangent to the hyperbola `x y=1`is a/anellipse(b) circlehyperbola (d) parabolaA. ellipseB. circleC. hyperbolaD. parabola |
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Answer» Correct Answer - C Let the point be (h, k). Then the equation of the chord of contact is `hx+ky=4.` Since `hx+ky=4` is tangent to xy = 1, `x((4-hx)/(k))=1` has two equal roots. Therefore, discriminant of `hx^(2)-4x+k=0` is 0. `therefore" "hk=4` Thus, the locus of (h, k) is xy = 4. |
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| 191. |
A tangent drawn to hyperbola `x^2/a^2-y^2/b^2 = 1` at `P(pi/6)` froms a triangle of area `3a^2` square units, with the coordinate axes, then the square of its eccentricity is (A) `15` (B)`24` (C) `17` (D) `14`A. 15B. 24C. 17D. 14 |
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Answer» Correct Answer - C `P(a sec.(pi)/(6),b tan.(pi)/(6))-=P((2a)/(sqrt3),(b)/(sqrt3))` Therefore, the equation of tangent at P is `(x)/(sqrt3a//2)-(y)/(sqrt3b)=1` `therefore" Area of triangle"=(1)/(2)xx(sqrt3a)/(2)xxsqrt3b=3a^(2)"(Given)"` `"or "(b)/(a)=4` `"or "e^(2)=1+(b^(2))/(a^(2))=17` |
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| 192. |
The locus a point `P(alpha,beta)` moving under the condition that the line `y=alphax+beta` is a tangent to the hyperbola `x^2/a^2-y^2/b^2=1` isA. a hyperbolaB. a parabolaC. a circleD. an ellipse |
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Answer» If `y=alpha x+beta` touches the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. ltbtgt then `beta^(2)=a^(2)alpha^(2)-b^(2)` [Putting `c=beta` and `m=alpha` in `c^(2)=a^(2)m^(2)-b^(2)`] Hence, locus of `P(alpha,beta)` is `y^(2)=a^(2)x^(2)-b^(2)` or , `a^(2)x^(2)-y^(2)=b^(2)` which represents a hyperbola. |
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| 193. |
A normal to the hyperbola `(x^2)/4-(y^2)/1=1`has equal intercepts on the positive x- and y-axis. If this normal touches the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`, then `a^2+b^2`is equal to5 (b)25 (c) 16(d) none of theseA. 5B. 25C. 16D. none of these |
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Answer» Correct Answer - D The equation of the normal to the hyperbola `(x^(2))/(4)-(y^(2))/(1)=1` at `(2 sec theta, tan theta)` is `2x cos theta+y cot theta=5`. The slope of the normal is `-2 sin theta=-1` `"or "sin theta=(1)/(2)or theta=(pi)/(6)` Y-intercept of the normal `=(5)/(cot theta)=(5)/(sqrt3)` As it touches the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` As it touches the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` we have `a^(2)+b^(2)=(25)/(3)` |
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| 194. |
Find the point on the hyperbola `x^2-9y^2=9`where the line `5x+12 y=9`touches it. |
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Answer» Solving `5x+12y=9 or y=(9-5x)/(12) and x^(2)-9y^(2)=9,` we have `x^(2)-9((9-5x)/(12))=9` `"or "x^(2)-(1)/(16)(9-5x)^(2)=9` `"or "16x^(2)-(25x^(2)-90x+81)=144` `"or "9x^(2)-(25x^(2)-90x+81)=144` `"or "9x^(2)-90x+225=0` `"or "x^(2)-10x+25=0` `"or "x=5` `rArr" "y=(9-25)/(12)=-(4)/(3)` So, point of contact is `(5,-(4)/(3))`. |
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| 195. |
If values of a, for which the line `y=ax+2sqrt(5)` touches the hyperbola `16x^2-9y^2 = 144` are the roots of the equation `x^2-(a_1+b_1)x-4=0`, then the values of `a_1+b_1` isA. 2B. 4C. zeroD. none of these |
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Answer» Correct Answer - C The equation of the hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1` The equation of the tangent is `y=mx+sqrt(9x^(2)-16)` `"or sqrt(9m^(2)-16)=2sqrt5` `"or "m= pm2` or a + b = sum of roots = 0 |
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| 196. |
A variable chord of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1,(b > a),`subtends a right angle at the center of the hyperbola if this chordtouches.a fixed circle concentric with the hyperbolaa fixed ellipse concentric with the hyperbolaa fixed hyperbola concentric with the hyperbolaa fixed parabola having vertex at (0, 0).A. a fixed circle concentric with the hyperbolaB. a fixed ellipse concentric with the hyperbolaC. a fixed hyperbola concentric with the hyperbolaD. a fixed parabola having vertex at (0, 0) |
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Answer» Correct Answer - A Let the variable chord be `x cos alpha+y sin alpha=p" (1)"` Let this chord intersect the hyperbola at A and B. Then the combined equation of OA and OB is given by `(x^(2))/(a^(2))-(y^(2))/(b^(2))=((x cos alpha+y sin alpha)/(p))^(2)` `x^(2)((1)/(a^(2))-(cos^(2)alpha)/(p^(2)))-y^(2)((1)/(b^(2))+(sin^(2)alpha)/(p^(2)))-(2 sin alpha cos alpha)/(p)xy=0` This chord subtends a right angle at the center. Therefore, `"Coefficient of " x^(2)+"Coefficient of "y^(2)=0` `"or "(1)/(a^(2))-(cos^(2)alpha)/(p^(2))-(1)/(b^(2))-(sin^(2) alpha)/(po)=0` `"or "(1)/(a^(2))-(1)/(b^(2))=(1)/(p^(2))` `"or "p^(2)=(a^(2)b^(2))/(b^(2)-a^(2))` Hence, p is constant, i.e., it touches the fixed circle. |
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| 197. |
A point `P`moves such that the chord of contact of the pair of tangents from `P`on the parabola `y^2=4a x`touches the rectangular hyperbola `x^2-y^2=c^2dot`Show that the locus of `P`is the ellipse `(x^2)/(c^2)+(y^2)/((2a)^2)=1.` |
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Answer» The equation of the chord of contact of parabola w.r.t. `P(x_(1),y_(1))` is given by `yy_(1)=2a(x+x_(1))" (1)"` Equation (1) touches the curve `x^(2)-y^(2)=c^(2)" (2)"` Using the condition of tangency on `y=(2ax)/(y_(1))+(2ax_(1))/(y_(1))` we get `(4a^(2)x_(1)^(2))/(y_(1))=c^(2)(4a^(2))/(y_(1)^(2))-c^(2)" "[because"(1) is tangent to"(x^(2))/(c^(2))-(y^(2))/(c^(2))=1]` `"or "4a^(2)x_(1)^(2)=4a^(2)c^(2)-c^(2)y_(1)^(2)` `"or "4a^(2)x_(1)^(2)+c^(2)y_(1)^(2)=4a^(2)c^(2)` Hence, the locus is `(x^(2))/(c^(2))+(y^(2))/((2a)^(2))=1` which is an ellipse. |
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| 198. |
If the line `y=2x+lambda` be a tangent to the hyperbola `36x^(2)-25y^(2)=3600`, then `lambda` is equal toA. `16`B. `-16`C. `+-16`D. none of these |
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Answer» The equation of the hyperbola is `(x^(2))/(100)-(y^(2))/(144)=1`. The line `y=2x+lambda` touches this hyperbola , if `lambda^(2)=100xx2^(2)-144` [Using `c^(2)=a^(2)m^(2)-b^(2)`] `implieslambda^(2)=256implieslambda`+-16` |
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| 199. |
The combined equation of the asymptotes of the hyperbola `2x^2 + 5xy + 2y^2 + 4x + 5y = 0` is -A. `2x^(2)+5xy+2y^(2)+4x+5y+2=0`B. `2x^(2)+5xy+2y^(2)+4x+5y-2=0`C. `2x^(2)+5xy+2y^(2)=0`D. none of these |
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Answer» Let the equation of asymptotes be `2x^(2)+5xy+2y^(2)+4x+5y+lambda=0`……`(i)` This equation represents a pair of straight lines. Therefore, `abc+2fgh-af^(2)-bg^(2)-ch^(2)=0` Here, `a=2`, `b=2`, `h=5//2`, `g=2`, `f=5//2` and `c=lambda` `:.4lambda+25-(25)/(2)-8-(25)/(4)lambda=0implies-(9lambda)/(4)+(9)/(2)=0implieslambda=2` Putting the value of `lambda` in `(i)`, we get `2x^(2)+5xy+2y^(2)+4x+5y+2=0` This is the equation of the asymptotes. |
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| 200. |
Find the equation of the hyperbola whose asymptotes are `3x=+-5y` and the vertices are `(+-5,0)`A. `3x^(2)-5y^(2)=25`B. `5x^(2)-3y^(2)=25`C. `9x^(2)-25y^(2)=225`D. `225x^(2)-9y^(2)=225` |
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Answer» The equation of the hyperbola having `3x+5y=0` and `3x-5y=0` as it asymptotes is ltrgt `(3x+5y)(3x-5y)+lambda=0`……..`(i)` It passes through `(5,0)` `implies225-0+lambda=0implieslambda=-225` Substituting the value of `lambda` in `(i)`, we get `9x^(2)-25y^(2)=225` as the equation of the hyperbola |
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