InterviewSolution
Saved Bookmarks
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.
| 201. |
If the line `y=3x+lambda` touches the hyperbola `9x^(2)-5y^(2)=45`, then `lambda`=A. `+-3sqrt(6)`B. `+-6`C. `+-3`D. `+-4` |
|
Answer» The equation of the hyperbola is `9x^(2)-5y^(2)=45implies(x^(2))/(5)-(y^(2))/(9)=1`……`(ii)` This of the form `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, where `a^(2)=5` and` b^(2)=9`. The line `y=3x+lambda` will touch the given hyperbola, if `lambda^(2)=5(3)^(2)-9` [Using : `c^(2)=a^(2)m^(2)-b^(2)`] `implieslambda^(2)=36implieslambda=+-6` |
|
| 202. |
The equation of tangent to hyperbola `4x^2-5y^2=20` which is parallel to `x-y=2` is (a) `x-y+3=0` (b) `x-y+1=0` (c) `x-y=0` (d) `x-y-3=0`A. `x-y-3=0`B. `x-y+9=0`C. `x-y+1=0`D. `x-y+7=0` |
|
Answer» Given equation of hyperbola is `4x^(2)-5y^(2)=20` which can be rewritten as ` rArr (x^(2))/(5)-(y^(2))/(4)=1` The line `x-y=2` has slope , `m=1` `therefore` Slope of tangent parallel to this line = 1 We know equation of tangent to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` having slope m is given by `y=mx pm sqrt(a^(2)m^(2)-b^(2))` Here, `a^(2)=5, b^(2)=4 and m=1` ` therefore` Required equation of tangent is `rArr y=xpm sqrt(5-4)` `rArr y=x pm 1 rArr x-ypm 1=0` |
|
| 203. |
In a hyperbola, the portion of the tangent intercepted between the asymptotes is bisected at the point of contact. Consider a hyperbola whose center is at the origin. A line `x+y=2` touches this hyperbola at P(1,1) and intersects the asymptotes at A and B such that AB = `6sqrt2` units. The equation of the tangent to the hyperbola at `(-1, 7//2)` isA. `5x+2y=2`B. `3x+2y=4`C. `3x+4y=11`D. none of these |
|
Answer» Correct Answer - B Let the equation of the hyperbola be `2x^(2)+2y^(2)+5xy+lambda=0` It passes through (1,1). Therefore, `2+2+5+lambda=0` `"or "lambda=-9` So, the hyperbola is `2x^(2)+2y^(2)+5xy=9` The equation of the tangent at `(-1,7//2)` is given by `2x(-1)+2y((7)/(2))+5(x(7//2)+(-1)y)/(2)=9` `"or "3x+2y=4` |
|
| 204. |
In a hyperbola, the portion of the tangent intercepted between the asymptotes is bisected at the point of contact. Consider a hyperbola whose center is at the origin. A line `x+y=2` touches this hyperbola at P(1,1) and intersects the asymptotes at A and B such that AB = `6sqrt2` units. The angle subtended by AB at the center of the hyperbola isA. `sin^(-11).(4)/(5)`B. `sin^(-1).(2)/(5)`C. `sin^(-1).(3)/(5)`D. none of these |
|
Answer» Correct Answer - C `m_(OA)=-(1)/(2),m_(OB)=-2` `tan theta=|(-1//2+2)/(1+1)|=(3)/(4)` `"or "sintheta=(3)/(5)` `"or "theta=sin^(-1)((3)/(5))` |
|
| 205. |
In a hyperbola, the portion of the tangent intercepted between the asymptotes is bisected at the point of contact. Consider a hyperbola whose center is at the origin. A line `x+y=2` touches this hyperbola at P(1,1) and intersects the asymptotes at A and B such that AB = `6sqrt2` units. The equation of the pair of asymptotes isA. `5xy+2x^(2)+2y^(2)=0`B. `3x^(2)+4y^(2)+6xy=0`C. `2x^(2)+2y^(2)-5xy=0`D. none of these |
|
Answer» Correct Answer - A The equation of tangent in parametric form is given by `(x-1)/(-1//sqrt2)=(y-1)/(1//sqrt2)= pm3sqrt2` `"or "A-=(4,-2), B-=(-2,4)` The equation of asymptotes (OA and OB) are given by `y+2=(-2)/(4)(x-4)` `"or "2y+x=0` `"and "y-4=(4)/(-2)(x+2)` `"or "2x+y=0` Hence, the combined equation of asymptotes is `(2x+y)(x+2y)=0` `"or "2x^(2)+2y^(2)+5xy=0` |
|
| 206. |
Find the equation of the hyperbola, whose axes are axes of coordinates and lengths of conjugate axis and latus rectum are 2 `(8)/(3)` respectively. |
|
Answer» Correct Answer - `16x^(2)-9y^(2) = 9` |
|
| 207. |
Find the centre, the length of latus rectum, the eccentricity, the coordinates of foci and the equations of the directrices of the hyperbola `((x+2)^(2))/(9) - ((y-1)^(2))/(16) = 1.` |
|
Answer» Correct Answer - (-2,1). `(sqrt(32))/(3), e=(5)/(3), (3,1) and (-7,1), 5(x+2) = pm9` |
|
| 208. |
Find the (i) centre, (ii) eccentricity, (iii) vertices and (iv) the equations of the directrices of the hyperbola `4x^(2) - 9y^(2) + 8x + 36y = 68.` |
|
Answer» Correct Answer - `pm (9)/(sqrt13)` |
|
| 209. |
If `e_1` is the eccentricity of the hyperbola `(y^(2))/(b^(2)) - (x^(2))/(a^(2)) = 1` then `e_(1)` =A. `sqrt(1+(a^(2))/(b^(2)))`B. `sqrt(1-(a^(2))/(b^(2)))`C. `sqrt(1+(b^(2))/(a^(2)))`D. `sqrt(1-(b^(2))/(a^(2)))` |
|
Answer» Correct Answer - C |
|
| 210. |
For the hyperbola `(x^2)/(cos^2alpha)-(y^2)/(sin^2alpha)=1`, which ofthe following remains constant when `alpha`varies?(1)eccentricity(2) directrix(3) abscissae of vertices (4) abscissaeof fociA. Abscissae of verticesB. Abscissae of fociC. EccentricityD. Directrix |
|
Answer» Given equation of hyperbola is `(x^(2))/(cos^(2)alpha)-(y^(2))/(sin^(2)alpha)=1 ` Here, `a^(2)=cos^(2)alpha and b^(2)=sin^(2)alpha` [i.e. comparing with standard equation `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`] We know that, foci `=(pm ae, 0)` where, `ae=sqrt(a^(2)+b^(2))=sqrt(cos^(2)alpha+sin^(2)alpha)=1` `rArr "Foci"=(pm 1, 0)` where, vertices are `(pm cos alpha,0).` Eccentricity, `ae=1 or e=(1)/(cos alpha)` Hence, foci remains constant with change in `alpha`. |
|
| 211. |
Find the equation of the hyperbola, the length ofwhose latusrectum is 8 and eccentricity is `3//sqrt(5)dot`A. `5x^(2)-4y^(2)=100`B. `4x^(2)-5y^(2)=100`C. `-4x^(2)+5y^(2)=100`D. `-5x^(2)+4y^(2)=100` |
|
Answer» Let the equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` ……..`(i)` We have, Length of the latusrectum `=8` `implies(2b^(2))/(a)=8` `impliesb^(2)=4a` `impliesa^(2)(e^(2)-1)=4a` `impliesa(e^(2)-1)=4impliesa((9)/(5)-1)=4impliesa=5` Putting `a=5 "in" b^(2)=4a`, we get `b^(2)=20`. Hence, the equation of the required hyperbola is `(x^(2))/(25)-(y^(2))/(20)=1` |
|
| 212. |
Find the asymptotes of the curve `x y-3y-2x=0`. |
|
Answer» Correct Answer - `x-3=0 and y-2=0` Since the equation of a hyperbola and its asymptotes differ in constant terms only, the pair of asymptotes is given by `xy-3y-2x+lambda=0" (1)"` where `lambda` is any constant such that it represents two straight lines. Hence, `abc+2fgh-af^(2)-bg^(2)-ch^(2)=0` `"or "0+2xx(-(3)/(2))xx(-1)xx((1)/(2))-0-0-lamda((1)/(2))^(2)=0` `"or "lambda=6` From (1), the asymptotes of the given hyperbola are given by `xy-3y-2x+6=0` `" "(y-2)(x-3)=0` Therefore, the asymptotes are `x-3=0 and y-2=0.` |
|
| 213. |
Find the angle between the asymptotes of the hyperbola `(x^2)/(16)-(y^2)/9=1`. |
|
Answer» Correct Answer - `tan^(-1)(24)/(7)` The angle between the asymptotes is given by `tan^(-1)((2ab)/(a^(2)-b^(2)))=tan^(-1){(2(4)(3))/(16-9)}` `=tan^(-1)((24)/(7))` |
|
| 214. |
Statement 1 : If from any point `P(x_1, y_1)`on the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=-1`, tangents are drawn to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1,`then the corresponding chord of contact lies on an other branch of thehyperbola `(x^2)/(a^2)-(y^2)/(b^2)=-1`Statement 2 : From any point outside the hyperbola,two tangents can be drawn to the hyperbola. |
|
Answer» Chord of contact of `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` w.r.t. point `P(x_(1),y_(1))` is `(x x_(1))/(a^(2))-(yy_(1))/(b^(2))=1" (1)"` Eq. (1) can be written as `(x(-x_(1)))/(a^(2))-(y(-y_(1)))/(b^(2))=-1`, which is tengent to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1` at point `(-x_(1),-y_(1))`. Obviously, points `(x_(1),y_(1)) and (-x_(1),-y_(1))` lie on the different branches of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1`. |
|
| 215. |
If the chords of contact of tangents from two points `(-4,2)` and `(2,1)` to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` are at right angle, then find then find the eccentricity of the hyperbola. |
|
Answer» Correct Answer - `sqrt((3)/(2))` Equation of chord of contact with respect to point `P(-4,2)` is `-(4x)/(a^(2))-(2y)/(b^(2))=1` and that with respect to point (2, 1) is `(2x)/(a^(2))-(y)/(b^(2))=1`. According to given condition, `(((4)/(a^(2)))/((-2)/(b^(2))))xx(((-2)/(a^(2)))/((-1)/(b^(2))))=-1` `rArr" "(b^(4))/(a^(4))=(1)/(4)` `rArr" "(b^(2))/(a^(2))=(1)/(2)` `therefore" "e=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+(1)/(2))=sqrt((3)/(2))` |
|
| 216. |
The locus of the poles of the chords of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` which subtend a right angle at its centre isA. `(x^(2))/(a^(4))+(y^(2))/(b^(4))=1`B. `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/(a^(2))+(1)/(b^(2))`C. `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/(a^(2))-(1)/(b^(2))`D. `(x^(2))/(a^(4))-(y^(2))/(b^(4))=(1)/(a^(2))-(1)/(b^(2))` |
|
Answer» Let `P(h,k)` be the pole of a chord of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. Then, the equation of the polar is `(hx)/(a^(2))-(ky)/(b^(2))=1`………`(i)` The combined equation of the lines joining the origin to the points of intersection of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` and the line `(i)` is `(x^(2))/(a^(2))-(y^(2))/(b^(2))=((hx)/(a^(2))-(ky)/(b^(2)))^(2)` Since the lines given by the above equation are at right angle. `:. "Coefficient of " x^(2)+ "Cofficient of" y^(2)=0` `implies(1)/(a^(2))-(h^(2))/(a^(4))-(1)/(b^(2))-(k^(2))/(b^(4))=0implies(h^(2))/(a^(4))+(k^(2))/(b^(4))=(1)/(a^(2))-(1)/(b^(2))` Hence, the locus of `P(h,k)` is `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/(a^(2))-(1)/(b^(2))` |
|
| 217. |
The circle `x^2+y^2-8x=0` and hyperbola `x^2/9-y^2/4=1` intersect at the points A and B Equation of the circle with AB as its diameter isA. `x^(2)+y^(2)-12x+24=0`B. `x^(2)+y^(2)+12x+24=0`C. `x^(2)+y^(2)+24x-12=0`D. `x^(2)+y^(2)-24x-12=0` |
|
Answer» The coordinates of any point on the given hyperbola are `(3sectheta,2tantheta)`. If lies on the circle `x^(2)+y^(2)-8x=0`, then `9sec^(2)theta+4tan^(2)theta-24sectheta=0` `implies13sec^(2)theta-24sectheta-4=0` `implies(13sectheta+2)(sectheta-2)=0` `impliessectheta=2`, `-(2)/(13)impliessectheta=2` `:.tantheta=+-sqrt(3)` So, the coordinates of `A` and `B` are `(6,2sqrt(3))` and `(6,-2sqrt(3))`. The equation of the circle with `AB` as its diameter is `(x-6)^(2)+(y^(2)+12)=0` or, `x^(2)+y^(2)-12x+24=0` |
|
| 218. |
Find the equations of the hyperbola satisfying the given conditions :Vertices `(+-2,0),f o c i(+-3,0)` |
|
Answer» Since the vertices of the given hyperbola are of the form `(+-a,0) ` it is a horizontal hyperbola . Let the required equation be `(x^(2))/(a^(2))=(y^(2))/(b^(2))=1.` then , its vertices are `(+-a,0)`. But , it is given that the vertices are `(+-2,0)`. `therefore a=2` Let its foci be `(+-c,0)` But , it is given that foci are `(+-3,0).` `therefore c=3.` `Now ,b^(2)=(c^(2)-a^(2))=(3^(2)-2^(2))=(9-4)=5.` Thus `,a^(2)=2^(2)=4 and b^(2)=5.` Hence , the required equation is `(x^(2))/(4)-(y^(2))/(b^(2))=1.` |
|
| 219. |
The circle `x^(2)+y^(2)-8x=0` and hyperbola `(x^(2))/(9)-(y^(2))/(4)=1` intersect at the points `A` and `B`. The equation of a common tangent with positive slope to the circle as well as to the hyperbola, isA. `2x-sqrt(5)y-20=0`B. `2x-sqrt(5)y+4=0`C. `3x-4y+8=0`D. `4x-3y+4=0` |
|
Answer» The equation of a tangent of slope `m` to the hyperbola `(x^(2))-(9)-(y^(2))/(4)=1` is `y=mx+sqrt(9m^(2)-4)` If it touches the circle `x^(2)+y^(2)-8x=0`, then `|(4m+sqrt(9m^(2)-4))/(sqrt(1+m^(2)))|=4` `implies495m^(4)+104m^(2)-400=0` `implies(5m^(2)-4)(99m^(2)+100)=0` `impliesm^(2)=(4)/(5)=(2)/(sqrt(5))` Substituting the value of `m` in `(i)`, we get `2x-sqrt(5)y+4=0` as the equation of the required common tangent |
|
| 220. |
Find the equation of the hyperbola with vertices at `(+- 6,0)` and foci at `(+-8,0).` |
| Answer» Correct Answer - `(x^(2))/(36)-(y^(2))/(28)=1` | |
| 221. |
The foci of a hyperbola are `(-5,18)` and `(10,20)` and it touches the `y`-axis . The length of its transverse axis, isA. `100`B. `sqrt(89)//2`C. `sqrt(89)`D. `sqrt(50)` |
|
Answer» Let `2a` and `2b` be respectively lengths of transverse and conjugate axes of the hyperbola and its eccentricity be `e`. Then, `2ae="Distance between foci"=2ae=17impliesae=(17)/(2)` We know that the product of lengths of perpendicular from two foci on any tangent to a hyperbola is `b^(2)`. Since given hyperbola touches `y`-axis i.e., `x=0` `:.b^(2)=50` `impliesa^(2)(e^(2)-1)=50implies(289)/(4)-a^(2)=50impliesa^(2)=(89)/(4)impliesa=(sqrt(89))/(2)` Hence, length of transverse axis `=2a=sqrt(89)`. |
|
| 222. |
Find the equations of the hyperbola satisfying the given conditions :Vertices `(0,+-5),f o c i (0,+-8)` |
| Answer» Correct Answer - `(y^(2))/(25)-(x^(2))/(39)=1` | |
| 223. |
Find the equation of the hyperbola whose foci are `(+- sqrt(29),0)` and the transverse axis is of the length 10. |
| Answer» Correct Answer - `(x^(2))/(25)-(y^(2))/(4)=1` | |