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.
| 1. |
If `4(x-sqrt2)^(2)+lambda(y-sqrt3)^(2)=45 and (x-sqrt2)^(2)-4(y-sqrt3)^(2)=5` cut orthogonally, then integral value of `lambda` is ________. |
|
Answer» Correct Answer - 9 Ellipse `((x-sqrt2)^(2))/(45//4)-((y-sqrt3)^(2))/(45//lambda)=1" cut orthogonally."` So, conics are confocal. `therefore" "(45)/(4)-(45)/(lambda)=5+(5)/(4)` `therefore" "lambda=9` |
|
| 2. |
Suppose the circle having equation `x^2+y^2=3`intersects the rectangular hyperbola `x y=1`at points `A ,B ,C ,a n dDdot`The equation `x^2+y^2-3+lambda(x y-1)=0,lambda in R ,`represents.a pair of lines through the origin for `lambda=-3`an ellipse through `A ,B ,C ,a n dD`for `lambda=-3`a parabola through `A , B , C ,a n dD`for `lambda=-3`a circle for any `lambda in R`A. a pair of lines through the origin for `lambda=-3`B. an ellipse through A, B, C and D for `lambda=-3`C. a parabola through A, B, C and D for `lambda=-3`D. a circle for any `lambda in R` |
|
Answer» Correct Answer - A For `lambda=-3`, the equation becomes `x^(2)+y^(2)-3xy=0` which represents a pair of lines through the origin. |
|
| 3. |
The equation of that chord of hyperbola `25x^(2)-16y = 400`, whose mid point is (5,3) isA. `115x - 117y = 17`B. `125x - 48y = 481`C. `127x + 33y = 341`D. `15x - 121y = 105` |
|
Answer» Correct Answer - B `S = 25 x^(2) - 16 y^(2) - 400 = 0` Equation of required chord is `S_(1) =T` (i) Here, `S_(1) = 25(5)^(2) - 16(3)^(2) - 400` `= 625 - 144 - 400 = 81` and `T = 25 xx_(1) - 16 yy_(1) - 400`, where `x_(1) = 5, y_(1) =3` `= 25(x)(5) - 16(y)(3) -400 - 125 x - 48y - 400` so from (i), required chord is `125x - 48y - 400 = 81` or `125x - 48y = 481` |
|
| 4. |
The equation of the chord joining two points `(x_(1),y_(1))` and `(x_(2),y_(2))` on the rectangular hyperbola `xy=c^(2)`, isA. `(x)/(x_(1)+x_(2))+(y)/(y_(1)+y_(2))=1`B. `(x)/(x_(1)-x_(2))+(y)/(y_(1)-y_(2))=1`C. `(x)/(y_(1)+y_(2))+(y)/(x_(1)+x_(2))=1`D. `(x)/(y_(1)-y_(2))+(y)/(x_(1)-x_(2))=1` |
|
Answer» Correct Answer - A The midpoint of the chord is `((x_(1)+x_(2))//2,(y_(1)+y_(2))//2)`. The equation of the chord in terms of its midpoint is `T=S_(1)` `"or "x((y_(1)+y_(2))/(2))+y((x_(1)+x_(2))/(2))-2c^(2)=((x_(1)+x_(2))/(2))((y_(1)+y_(2))/(2))-2c^(2)` `"or "x(y_(1)+y_(2))+y(x_(1)+x_(2))=(x_(1)+x_(2))(y_(1)+y_(2))` `"or "(x)/(x_(1)+x_(2))+(y)/(y_(1)+y_(2))=1` |
|
| 5. |
If the chord joining the points `(a sectheta_1, b tantheta_1)` and `(a sectheta_2, b tantheta_2)` on the hyperbola `x^2/a^2-y^2/b^2=1` is a focal chord, then prove that `tan(theta_1/2)tan(theta_2/2)+(ke-1)/(ke+1)=0`, where `k=+-1`A. `(1-e)/(1+e)`B. `(e-1)/(e+1)`C. `(e+1)/(e-1)`D. `(1+e)/(1-e)` |
|
Answer» The equation of the chord joining `(a sectheta_(1), b tantheta_(1))` and `(a sec theta_(2), b tan theta_(2))` is `(x)/(a)cos((theta_(1)-theta_(2))/(2))-(y)/(b)sin((theta_(1)+theta_(2))/(2))=cos((theta_(1)+theta_(2))/(2))` If it passes through the focus `(ae,0)` then `e cos ((theta_(1)-theta_(2))/(2))=cos((theta_(1)+theta_(2))/(2))` `implies(cos((theta_(1)-theta_(2))/(2)))/(cos((theta_(1)+theta_(2))/(2)))=(1)/(e)` `implies(cos((theta_(1)-theta_(2))/(2))+cos((theta_(1)+theta_(2))/(2)))/(cos((theta_(1)+theta_(2))/(2))-cos((theta_(1)+theta_(2))/(2)))=(1+e)/(1-e)` `impliescot.(theta_(1))/(2)cot.(theta_(2))/(2)=(1+e)/(1-e)impliestan.(theta_(1))/(2)tan.(theta_(2))/(2)=(1-e)/(1+e)` |
|
| 6. |
If a chord joining `P(a sec theta, a tan theta), Q(a sec alpha, a tan alpha)` on the hyperbola `x^(2)-y^(2) =a^(2)` is the normal at P, then `tan alpha =`A. `tan theta (4 sec^(2) theta+1)`B. `tan theta (4 sec^(2) theta -1)`C. `tan theta (2 sec^(2) theta -1)`D. `tan theta (1-2 sec^(2) theta)` |
|
Answer» Correct Answer - B Slope of chord joining P and Q = slope of normal at P `:. (tan alpha - tan theta)/(sec alpha - sec theta) =- (tan theta)/(sec theta)` `:. tan alpha - tan alpha =- k tan theta` and `sec alpha - sec theta = k sec theta (1+k) sec theta = sec alpha` (1) `:. (1-k) tan theta = tan alpha` (2) `[(1+k)sec theta]^(2) - [(1-k)tan theta]^(2) = sec^(2) alpha - tan^(2) alpha =1` `rArr k =- 2 (sec^(2) theta + tan^(2) theta) =- 4 sec^(2) theta +2` From (2), `tan alpha = tan theta (1+4 sec^(2) theta -2) = tan theta (4 sec^(2) theta -1)` |
|
| 7. |
Prove that any hyperbola and its conjugate hyperbola cannot have common normal. |
|
Answer» Consider hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1.` Equation of normal to hyperbola at point `P(a sec theta, b tan theta)` is `ax cos theta+by cot theta=a^(2)+b^(2)" (1)"` Equation of normal to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1` at point `Q( a tan phi, b sec phi)` is `ax cot phi+"by" cos phi=a^(2)+b^(2)" (2)"` If Eqs. (1) and (2) represent the same straight line, then `(cot phi)/(cos theta)=(cos phi)/(cot theta)=1` `rArr" "tan phi = sec theta and sec phi = tan theta` `rArr" "sec^(2)phi-tan^(2)phi=tan^(2)theta-sec^(2)theta=-1,` which is not possible. Thus, hyperbola and its conjugate hyperbola cannot have common normal. |
|
| 8. |
The number of normal (s) of a rectangular hyperbola which can touch its conjugate is equal to |
|
Answer» Correct Answer - C Normal to hyperbola `xy = c^(2)` at `(ct, (c )/(t))` is `y - (c )/(t) = t^(2) (x-ct)` Solving with `xy =- c^(2)`, we get `rArr x {(c )/(t) + t^(2) (x-ct)} +c^(2) =0` `rArr t^(2) x^(2) + ((c )/(t)-ct^(3)) x +c^(2) =0` Since line touches the curve, above equation has equal roots `:. Delta = 0, ((1)/(t)-t^(3))^(2) - 4t^(2) =0` `rArr (1-t^(4))^(2) - 4t^(4) =0` `rArr t^(2) = (2=- sqrt(8))/(2)` or `(-2 +- sqrt(8))/(2)` `rArr t^(2) =1 + sqrt(2)` or `-1 + sqrt(2)` Thus four such values of t are possible |
|
| 9. |
Normal are drawn to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`at point `theta_1a n dtheta_2`meeting the conjugate axis at `G_1a n dG_2,`respectively. If `theta_1+theta_2=pi/2,`prove that `C G_1dotC G_2=(a^2e^4)/(e^2-1)`, where `C`is the center of the hyperbola and `e`is the eccentricity. |
|
Answer» The normal at point `P(a sec theta_(1), b tan theta_(2))` is `ax cos theta_(1)+by cot theta_(1)=(a^(2)+b^(2))` It meets the conjugate axis at `G_(1)(0,(a^(2)+b^(2))/(b)tan theta_(1))`. The normal at point `Q(a sec theta_(2), b tan theta_(2))` is `axcos theta_(2)+" by " cot theta_(2)=(a^(2)+b^(2))` It meets the conjugate axis at `G_(1)(0,(a^(2)+b^(2))/(b)tan theta_(2)).` Therefore, `CG_(1)*CG_(2)=((a^(2+b^(2)))^(2))/(b^(2))tantheta_(1)tantheta_(2)` `=((a^(2)+b^(2)))/(b^(2))" "(becausetheta_(1)+theta_(2)=(pi)/(2))` `=(a^(4)(1+(b^(2))/(a^(2))))/(b^(2))` `=(a^(2)e^(4))/(e^(2)-1)` |
|
| 10. |
There exist two points `P` and `Q` on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` such that `PObotOQ`, where `O` is the origin, then the number of points in the `xy`-plane from where pair of perpendicular tangents can be drawn to the hyperbola , isA. `0`B. `1`C. `2`D. infinite |
|
Answer» Let `P(a sectheta_(1),btantheta_(1))` and `Q(a sectheta_(2),btantheta_(2))` be two points on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` such that `OPbotOQ` `implies(b)/(a)(tantheta_(1))/(sectheta_(1))xx(b)/(a)(tantheta_(2))/(sectheta_(2))=-1` `impliessintheta_(1)sintheta_(2)=-(a^(2))/(b^(2))` `impliesa^(2) lt b^(2)` The point on `xy`-plane from where perpendicular tangents are drawn to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` lie on the director circle `x^(2)-y^(2)=a^(2)-b^(2)`. For `a^(2) lt b^(2)` the director circle does not exist.So, there is no point in `xy`-plane. |
|
| 11. |
Statement-`1` : If the foci of a hyperbola are at `(4,1)` and `(-6,1)` and eccentricity is `(5)/(4)`, then the length of its transverse axis is `4`. Statement-`2` : Distance between the foci of a hyperbola is equal to the product of its eccentricity and length of the transverse axis.A. `1`B. `2`C. `3`D. `4` |
|
Answer» Distance between two foci of the hyperbola `=2ae` `:. 4+6=2aeimplies10=2xx(5)/(4)xxaimpliesa=4`. Hence, length of transverse axis `=8`. So, statement-`1` is not true. Distance between foci `=2ae=` Transverse axis `xx` Eccentricity. So, statement-`2` is not true. |
|
| 12. |
A hyperbola whose transverse axis is along the major axis of the conic,`x^2/3+y^2/4= 4` and has vertices atthe foci of this conic. If the eccentricity of the hyperbola is `3/2`, then which of the following points does NOT lie on it?A. `(sqrt(5),2sqrt(2))`B. `(sqrt(5),2sqrt(3))`C. `(0,2)`D. `(sqrt(10),2sqrt(3))` |
|
Answer» The ellipse `(x^(2))/(3)+(y^(2))/(4)=4`, or `(x^(2))/(12)+(y^(2))/(16)=1` has its major axis along `y`-axis and eccentricity `e=sqrt(1-(12)/(16))=(1)/(2)`. So, coordinates of its foci are `(0,2)` and `(0,-2)` . The vertices of the hyperbola are at `(0,2)` and `(0,-2)` .So, the length of its transverse axis is `2b=4`. Let the length of its conjugate axis be `2a`. Then, `a^(2)=b^(2)(e^(2)-1)impliesa^(2)=4((9)/(4)-1)=5` So, the equation of the hyperbola is `(x^(2))/(5)-(y^(2))/(4)=-1` Clearly, `(5,2sqrt(3))` does not lie on this hyperbola. |
|
| 13. |
If the curves `x^(2)-y^(2)=4` and `xy = sqrt(5)` intersect at points A and B, then the possible number of points (s) C on the curve `x^(2)-y^(2) =4` such that triangle ABC is equilateral is |
|
Answer» Correct Answer - A A and B are `(sqrt(5),1)` and `(-sqrt(5),-1)`. Let C be `(2 sec theta, 2 tan theta)` `O(0,0)` is the mid point of Ab Slope of `OC = sin theta` and slope of `AB = (1)/(sqrt(5))` Since `OC _|_AB` So, `sin theta =- sqrt(5)` which is impossible. |
|
| 14. |
The chord of contact of a point `P`w.r.t a hyperbola and its auxiliary circle are at right angle. Then thepoint `P`lies onconjugate hyperbolaone of the directrixone of the asymptotes(d) none of theseA. conjugate hyperbolaB. one of the directrixC. asymptotesD. none of these |
|
Answer» Correct Answer - C Let `P(h,k)` be any point. The chord of contact of P w.r.t. the hyperbola is `(hx)/(a^(2))-(ky)/(b^(2))=1" (1)"` The chord of contact of P w.r.t. the auxxiliary circle is `hx+ky=a^(2)" (2)"` Now, `(h)/(a^(2))xx(b^(2))/(k)xx(-(h)/(k))=-1` `"or "(h^(2))/(a^(2))-(k^(2))/(b^(2))=0` Therefore, P lies on one of the asymptotes. |
|
| 15. |
The point `(3tan (theta +60^(@)),2 tan(theta +30^(@)))` lies on the conic, then its centre is `(theta` is the parameter)A. `(-3sqrt(3),2sqrt(3))`B. `(3sqrt(3),-2sqrt(3))`C. `(-3sqrt(3),-2sqrt(3))`D. (0,0) |
|
Answer» Correct Answer - A Let `(3 tan (theta + 60^(@)),2 tan (theta + 30^(@)) -= (h,k)` `:. tan (theta + 60^(@)) = (h)/(3)` (1) and `tan (theta + 30^(@)) = (k)/(2)` (2) `tan 30^(@) = tan [(theta + 60^(@))- (theta + 30^(@))]` `rArr (1)/(sqrt(3)) = (tan (theta+60^(@))-tan(theta+30^(@)))/(1+tan (theta+60^(@))tan (theta+30^(@)))` `rArr (1)/(sqrt(3)) =((x)/(3)-(y)/(2))/(1+(xy)/(6))` `rArr xy - 2sqrt(3)x + 3sqrt(3)y + 6 =0` `rArr (x+3sqrt(3)) (y-2sqrt(3)) + 24 =0` `rArr` center is `(-3sqrt(2),2sqrt(3))` |
|
| 16. |
On which curve does the perpendicular tangents drawn to the hyperbola `(x^(2))/(25)-(y^(2))/(16)=1` intersect? |
|
Answer» The locus of the point of intersection of prependicular tangents to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is the director circle given by `x^(2)+y^(2)=a^(2)-b^(2)` Hence, the perpendicular tangents drawn to `(x^(2))/(25)-(y^(2))/(16)=1` intersect on the curve `x^(2)+y^(2)=25-16=9` |
|
| 17. |
The number of points on the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=3`from which mutually perpendicular tangents can be drawn to the circle `x^2+y^2=a^2`is/are0 (b)2 (c) 3(d) 4 |
|
Answer» Correct Answer - A Director circle of the circle `x^(2)+y^(2)=a^(2)" is "x^(2)+y^(2)=2a^(2)`. The semi-transverse axis is `sqrt3a`. The radius of the circle is `sqrt2a`. Hence, director circle and hyperbola do not intersect. |
|
| 18. |
Find the equation of pair of tangents drawn from point (4, 3) to the hyperbola `(x^(2))/(16)-(y^(2))/(9)=1`. Also, find the angle between the tangents. |
|
Answer» Equation of pair of tengents is `T^(2)=SS_(1)`. `therefore" "((4x)/(16)-(3y)/(9)-1)^(2)=((x^(2))/(16)-(y^(2))/(9)-1)(-1)` `rArr" "(x^(2))/(16)+(y^(2))/(9)+1-(xy)/(6)-(x)/(2)+(2y)/(3)=-(x^(2))/(16)+(y^(2))/(9)+1` `rArr" "(x^(2))/(8)-(xy)/(6)-(x)/(2)+(2y)/(3)=0` `rArr 3x^(2)-4xy-12x+16y=0,` which is requird equation of pair of tangents. Comparing it with standard second-degree equation, we have `a=3,b=0 and h=-2` `therefore` Angle between tangents, `theta=tan^(-1).(2sqrt(h^(2)-ab))/(|a+b|)` `=tan^(-1).(2sqrt((-2)^(2)-(3)(0)))/(|3+0|)` `=tan^(-1).(4)/(3)` |
|
| 19. |
If `P(alpha, beta)`, the point of intersection of the ellipse `x^2/a^2+y^2/a^2(1-e^2)=1` and hyperbola `x^2/a^2-y^2/(a^2(E^2-1)=1/4`is equidistant from the foci of the curvesall lying in the right of y-axis thenA. `2 alpha =a (2e +E)`B. `a- ealpha = E alpha -alpha//2`C. `E =(sqrt(e^(2)+24)-3e)/(2)`D. `E=(sqrt(e^(2)+12)-3e)/(2)` |
|
Answer» Correct Answer - B::C Focus of ellipse lying to the right of the y-axis is `S_(1) (ae,0)` Focus of hyperbola lying to the right of the y-axis is `S_(2)(aE//2,0)` Now `P(alpha, beta)` is equidistance from `S_(1)` and `S_(2)` `:. S_(1)P = S_(2)P rArr a - e alpha = E alpha -((a)/(2))`. (1) Also P lies on perpendicular bisector of `S_(1)S_(2)` `:. alpha = (ae+(a)/(2)E)/(2)` From (1) and (2), `E^(2) + 3eE + (2e^(2)-6) =0` `rArr E = (sqrt(e^(2)+24)-3e)/(2)` |
|
| 20. |
If tangents drawn from the point `(a ,2)`to the hyperbola `(x^2)/(16)-(y^2)/9=1`are perpendicular, then the value of `a^2`is _____ |
|
Answer» Correct Answer - 3 Since tangents drawn from the point `A(a,2)` are perpendiular, A must lie on the director circle `x^(2)+y^(2)=7`. Putting y = 2, we get `x^(2)=a^(2)=3.` |
|
| 21. |
If `P(a sec alpha,b tan alpha)` and `Q(a secbeta, b tan beta)` are two points on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` such that `alpha-beta=2theta` (a constant), then `PQ` touches the hyperbolaA. `(x^(2))/(a^(2)sec^(2)theta)-(y^(2))/(b^(2))=1`B. `(x^(2))/(a^(2))-(y^(2))/(b^(2)sec^(2)theta)=1`C. `(x^(2))/(a^(2))-(y^(2))/(b^(2))=cos^(2)theta`D. none of these |
|
Answer» The equation of chord `PQ` is `(x)/(a)cos((alpha-beta)/(2))-(y)/(b)sin((alpha+beta)/(2))=cos((alpha+beta)/(2))` `implies(x)/(a)costheta-(y)/(b)sin((alpha+beta)/(2))=cos((alpha+beta)/(2))` `implies(x)/(asectheta)sec((alpha+beta)/(2))-(y)/(b)tan((alpha+beta)/(2))=1` Clearly, it touches the hyperbola `(x^(2))/(a^(2)sec^(2)theta)-(y^(2))/(b^(2))=1` |
|
| 22. |
Find the equation of the hyperbola, whose axes are axes of coordinates and distance between the foci is 10 and length of conjugate axis is 6. |
|
Answer» Correct Answer - `9x^(2)-16y^(2) = 144` |
|
| 23. |
Find the equation of the asymptotes of the hyperbola `3x^2+10 x y+9y^2+14 x+22 y+7=0` |
|
Answer» Since the equation of hyperbola and the combined equation of its asymptotes differ by a constant, the equations of the asymptotes should be `3x^(2)+10xy+8y^(2)+14x+22y+lambda=0" (1)"` Now, `lambda` is to be chosen so that (1) represents a pair of straight lines. Comparing (1) with `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0" (2)"` we have `a=3,b=8,h=5,g=7,f=11, c=lambda` We know that (2) represents a pair of straight lines if `abc+2hgf-af^(2)-bg^(2)-ch^(2)=0` `"or "3xx8xxlambda+2xx7xx11xx5-3xx121-8xx49-lambdaxx25=0` `"or "lambda=15` Hence, the combined equation of the asymptotes is `3x^(2)+10xy+8y^(2)+14x+22y+15=0`. |
|
| 24. |
If radii of director circles of `x^2/a^2+y^2/b^2=1 and x^2/a^2-y^2/b^2=1` are `2r and r` respectively, let `e_E and e_H` are the eccentricities of ellipse and hyperbola respectively, then |
|
Answer» Correct Answer - 6 Equation of director cicles of ellipse and hyperbola are, respectively, `x^(2)+y^(2)=a^(2)+b^(2)` `and x^()+y^(2)=a^(2)-b^(2)` According to the question, we have `a^(2)+b^(2)=4r^(2)" (1)"` `a^(2)-b^(2)=r^(2)" (2)"` Solving Eqs. (1) and (2), we get `a^(2)=(5r^(2))/(2),b^(2)=(3r^(2))/(2)` `e_(1)^(2)=1-(b^(2))/(a^(2))` `=1-(3)/(5)=(2)/(5)` `e_(2)^(2)=1+(b^(2))/(a^(2))` `=1+(3)/(5)=(8)/(5)` So, `4e_(2)^(2)-e_(1)^(2)=4xx(8)/(5)-(2)/(5)=6` |
|
| 25. |
If radii of director circles of `x^2/a^2+y^2/b^2=1 and x^2/a^2-y^2/b^2=1` are `2r and r` respectively, let `e_E and e_H` are the eccentricities of ellipse and hyperbola respectively, thenA. `2e_(h)^(2)-e_(e)^(2)=6`B. `e_(e)^(2)-4e_(h)^(2)=6`C. `4e_(h)^(2)-e_(e)^(2)=6`D. none of these |
|
Answer» The equation of the director circles of ellipse and hyperbola are `x^(2)+y^(2)=a^(2)+b^(2)` and `x^(2)+y^(2)=a^(2)-b_(1)^(2)` respectively. `:.2r=sqrt(a^(2)+b^(2))` and `r=sqrt(a^(2)-b_(1)^(2))` `=2sqrt(a^(2)-b_(1)^(2))=sqrt(a^(2)+b^(2))` `implies4(a_(1)^(2)-b_(1)^(2))=a^(2)+b^(2)` `implies4(1-(b_(1)^(2))/(a^(2)))=(1+(b^(2))/(a^(2)))` `implies4{1-(e_(h)^(2)-1){=1+(1-e_(e)^(2))` `implies8-4e_(h)^(2)=2-e_(e)^(2)implies4e_(h)^(2)-e_(e)^(2)=6` |
|
| 26. |
Find the equation of the hyperbola, whose axes are axes of coordinates and coordinates of foci are `(pm(5)/(2),0)` and the length of latus rectum is `(9)/(4)`. |
|
Answer» Correct Answer - `9x^(2)-16y^(2) = 36` |
|
| 27. |
Find the equation of normal to the hyperbola `x^2-9y^2=7`at point (4, 1). |
|
Answer» Differentiating the equation of hyperbola `x^(2)-9y^(2)=7a` w.r.t. x, we get `2x-18y(dy)/(dx)=0` `"or "(dy)/(dx)=(x)/(9y)` slope of normal at any point one the curve is `-(dx)/(dy)=-(9y)/(x).` Therefore, the slope of noraml at point (4, 1) is `(-(dx)/(dy))_(("4,1"))=-(9)/(4)` Therefore, the equation of normal at point (4, 1) is `y-1=-(9)/(4)(x-4)` `"or "9x+4y=40` Alternative method, For hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, equation of normal at point `P(x_(a),y_(1))` is `(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))==a^(2)+b^(2)` So, for given hyperbola `(x^(2))/(7)-(y^(2))/(7//9)=1`, equation of normal at point P(4, 1) is `(7x)/(4)+((7//9)y)/(1)=7+(7)/(9)` `rArr" "9x+4y=40` |
|
| 28. |
The normal to a curve at `P(x, y)` meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a (1) ellipse (2) parabola (3) circle (4) hyperbola |
|
Answer» `(Y-y)=-dx/(dy)(X-x)` `0-y=-(dx)/(dy)(G-x)` `G=y(dy)/(dx)+x` `|G|=2|x|` `|x+(y)dy/(dx)|=2|x|` `x+(y)dy/(dx)=2x or x+y(dy)/(dx)=-2x` `y(dy)/(dx)=x or y(dy)/(dx)=-x` `inty dy=intxdx or intydy=int-3xdx` `y^2/2=x^2/2+c or y^2/2=-3x^2/2+c` curves are `hyperbola or ellipse` |
|
| 29. |
Let `S={(x,y) in R^(2):(y^(2))/(1+r)-(x^(2))/(1-r)=1}`, where `r ne pm 1`. Then S represents:A. a hyperbola whose eccentricity is `(2)/(sqrt(1-r)),` when `0 lt r lt 1.`B. a hyperbola whose eccentricity is `(2)/(sqrt(r+1)),` when `0 lt r lt 1.`C. an ellipse whose eccentricity is `sqrt((2)/(r+1)) " when "r gt 1`.D. an ellipse whose eccentricity is `(1)/(sqrt(r+1)) " when "r gt 1`. |
|
Answer» Given, `S={(x,y) in R^(2): (y^(2))/(1+r)-(x^(2))/(1-r)=1}` `={(x,y) in R^(2): (y^(2))/(1+r)+(x^(2))/(r-1)=1}` For `r gt 1,(y^(2))/(1+r)+(x^(2))/(r-1)=1`, represents a vertical ellipse. `" "[ because " for " r gt 1, r-1 lt r+1 and r-1 gt 0]` Now, eccentricity (e)`=sqrt(1-(r-1)/(r+1))` `[ because " For " (x^(2))/(a^(2))-(y^(2))/(b^(2))=1, a lt b, e -sqrt(1-(a^(2))/(b^(2)))]` `=sqrt(((r+1)-(r-1))/(r+1))` `=sqrt((2)/(r+1))` |
|
| 30. |
From a point `P(1,2)`, two tangents are drawn to a hyperbola `H`in which one tangent is drawn to each arm of the hyperbola. If theequations of the asymptotes of hyperbola `H`are `sqrt(3)x-y+5=0`and `sqrt(3)x+y-1=0`, then the eccentricity of `H`is2 (b) `2/(sqrt(3))`(c) `sqrt(2)`(d) `sqrt(3)` |
|
Answer» We know that if angle between asymptotes is `theta,` then eccentricity of hyparbola is `sec.(theta)/(2)`. A cute angle between asmytotes `sqrt3x-y+5=0` and `sqrt3x+y-1=0" is "(pi)/(3)`. Let `L_(1)(x,y)=sqrt3x-y+5 and L_(2)(x,y)=sqrt3x+y-1`. `therefore" "L_(1)(0,0)=5 and L_(2)(0,0)=-1` Also, `a_(1)a_(2)+b_(1)b_(2)=(sqrt3)(sqrt3)+(-1)(1)=2gt0`. Thus, origin lies in acute angle. Now, `L_(1)(1,2)gt0 and L_(2)(1,2)gt0` So, (1,2) lies in obtuse angle formed by asymptotes. Since tangents drawn from point (1, 2) are to different branches of hyperbola, brnaches of hyperbola lie in acute angles formed by asymptotes. Therefore, eccentricity of hyperbola is `e=sec.(theta)/(2)=sec.(pi)/(6)=(2)/(sqrt3).` |
|
| 31. |
The foci of the ellipse `(x^(2))/(25) + (y^(2))/(9) = 1` coincide with the hyperbola. It e = 2 for hyperbola, find the equation of the hyperbola. |
|
Answer» Correct Answer - `3x^(2) - y^(2) = 12` |
|
| 32. |
Find all the aspects of hyperbola `16x^(2)-3y^(2)-32x+12y-44=0.` |
|
Answer» Correct Answer - `e=sqrt(19//3),"Vertices"-=(1pmsqrt3,2),"Foci"-=(1pmsqrt(19),2), "Directrix:"x-1=pm3//sqrt(19)` We have, `16(x^(2)-2x)-3(y^(2)-4y)=44` `"or "16(x-1)^(2)-3(y-2)^(2)=48` `"or "((x-1)^(2))/(3)-((y-2)^(2))/(16)=1` Clearly, transverse axis is horizontal, having equation `y-2=0.` Conjugate axis is `x-1=0 or x=1.` Centre is (1,2). Here, `a=sqrt3` and b = 4. Thus, length of transverse axis is `2sqrt3` and that of conjugate axis is 8. `"Eccentricity, e"=sqrt(1+(16)/(3))=sqrt((19)/(3))` Vertices lie at distance a from centre on transverse axis. Therefore, vertices are `(1pm sqrt3, 2)` Foci lie at distance ae from centre on transverse axis. Therefore, foci are `(1pm sqrt(19),2)`. Equation of directrices are `x-1= pm(1)/(e)` `"or "x-1=pm(3)/(sqrt(19)` |
|
| 33. |
The equation `px^(2) + 2qxy + ry^(2) + 2sx + 2ty + u = 0` (where `p ne q ne r ` and p,q,r,s,t and u are constants ) represents-A. an ellipse if p,q,r, in H.P.B. a parabola if p,q,r, in G.P.C. a hyperbola if p,q,r in A.P.D. A,B,C will be true if `|(p,q,r),(q,r,t),(s,t,u)| ne 0` |
|
Answer» Correct Answer - A,B,C,D |
|
| 34. |
Show that the locus represented by `x=(1)/(2)a(t+(1)/(t)),y=(1)/(2)a(t-(1)/(t))` is a rectangular hyperbola. |
|
Answer» Let point P be `(a sec theta, b tan theta)`. Equation of tangent at point P is `(x)/(a)sec theta-(y)/(b)tan theta=1` Equation of asymptotes are `y=pm(b)/(a)x.` Solving asymptotes with tangent, we get `Q-=((a)/(sectheta-tantheta),(b)/(sectheta-tantheta))` `"and "R-=((a)/(sectheta+tantheta),(-b)/(sectheta+tantheta))` `therefore" Area of triangle CQR"=(1)/(2)||(0,0),((a)/(sectheta-tantheta),(b)/(sectheta-tantheta)),((a)/(sectheta+tantheta),(-b)/(sectheta+tantheta)),(0,0)||` `=(1)/(2)|-(a)/(sectheta-tantheta).(b)/(sectheta+tantheta)-(a)/(sectheta-tantheta).(b)/(sectheta+tantheta)|` = ab, which is constant. Also, midpoint of QR is `(((1)/(sectheta-tantheta)+(1)/(sectheta+tantheta))/(2),((b)/(sectheta-tantheta)-(b)/(sectheta+tantheta))/(2))` `-=(a sec theta, b tan theta)`, which is point P. |
|
| 35. |
Explain what geometrical diagrams are represented by the equatio `(x^(2))/(a^(2)) + (y^(2))/(a^(2)(1-e^(2))) = 1`, where a and e are constants. |
|
Answer» Correct Answer - An ellipse when `0 lt e lt 1`, a circle when e =0 and a hyperbola when e >1. |
|
| 36. |
The hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2)) = 1` passes through the point (-3,2) and its eccentricity is `sqrt((5)/(3))`, find the length of its latus rectum. |
|
Answer» Correct Answer - `(4)/(3)sqrt(3)` |
|
| 37. |
Find the equation of the hyperbola, whose axes are axes of coordinates and coordinates of foci are (5,0),(-5,0) and the eccentricity is `(5)/(4)`. |
|
Answer» Correct Answer - `9x^(2)-16y^(2) = 144` |
|
| 38. |
The hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2)) = 1` passes through the point of intersection of the lines 7x + 13y - 87 = 0 and 5x - 8y + 7 = 0 and its latus rectum is `(32sqrt(2))/(5)` Find a and b. |
|
Answer» Correct Answer - `a = (5)/(sqrt(2)), b = 4` |
|
| 39. |
For what value of m the hyperbola `3x^(2)-my^(2) = 48` will pass through the poin (8,-6) ? Find its eccentricity and the length of latus rectum. |
|
Answer» Correct Answer - m = 4, `e=(sqrt(7))/(2)` and length of latus rectum = 6 unit |
|
| 40. |
Show that for all values of t the point x = a. `(1 + t^(2))/(1-t^(2)), y = (2at)/(1-t^(2))`lies on a fixed hyperbola. What is the value of the eccentricity of the hyperbola? |
|
Answer» Correct Answer - `e=sqrt(2)` |
|
| 41. |
P is a veriable poin on the hyperbola `x^(2) - y^(2) =16` and A (8,0) is a fixed point . Show that the locus of the middle point of the line segment `bar(AP)` is another rectangular hyperbola. |
|
Answer» Correct Answer - `3y^(2)-x^(2) = 12` |
|
| 42. |
The locus of the foot of perpendicular from my focus of a hyperbola upon any tangent to the hyperbola is the auxiliary circle of the hyperbola. Consider the foci of a hyperbola as `(-3, -2)` and (5,6) and the foot of perpendicular from the focus (5, 6) upon a tangent to the hyperbola as (2, 5). The conjugate axis of the hyperbola isA. `4sqrt(11)`B. `2sqrt(11)`C. `4sqrt(22)`D. `2sqrt(22)` |
|
Answer» Correct Answer - D Centre `-=(1,2)` Radius of auxiliary circle `=a =sqrt((2-1)^(2)+(5-2)^(2))=sqrt(10)` `2ae=sqrt(8^(2)+8^(2))=8sqrt2 or e=(4)/(sqrt5)` `b^(2)=a^(2)e^(2)-a^(2)=32-10=22` `"or "2b=2sqrt(22)` |
|
| 43. |
The foci of a hyperbola coincide with the foci of the ellipse `(x^(2))/(25)+(y^(2))/(9)=1`. If the eccentricity of the hyperbola is `2`, then the equation of the tangent of this hyperbola passing through the point `(4,6)` isA. `2x-y-2=0`B. `3x-2y=0`C. `2x-3y+10=0`D. `x-2y+8=0` |
|
Answer» For the ellipse `(x^(2))/(25)+(y^(2))/(9)=1`, we have `a=5`, `b=3`. `:.e=sqrt(1-(9)/(25))=(4)/(5)` So, the coordinates of the foci are `(+-4,0)`. These are also foci of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` whose eccentricity is `2`. `:.ae=4impliesa=2` Now, `b^(2)=a^(2)(e^(2)-1)impliesb^(2)=4(4-1)=12` So, the equation of the hyperbola is `(x^(2))/(4)-(y^(2))/(12)=1`. Clearly , point `(4,6)` lies on it. The equation of tangent to this hyperbola at `(4,6)` is `(4x)/(4)-(6y)/(12)=1` or, `2x-y=2` |
|
| 44. |
Prove that the locus of the point of intersection of the tangents atthe ends of the normal chords of the hyperbola `x^2-y^2=a^2`is `a^2(y^2-x^2)=4x^2y^2dot` |
|
Answer» Normal at a point `(a sec theta, a tan theta)` is `x cos theta+y cot theta=2a` If `P(x_(1),y_(1))` is the point of intersection of the tangents at the ends of normal chord (1), then (1) must be the chord of contact of P(h, k) whose equation is given by `hx-ky=a^(2)" (2)"` Comparing (1) and (2) and eliminating `theta`, we get `(a^(2))/(4h^(2))-(a^(2))/(4k^(2))=1` Hence, the locus is `(1)/(x^(2))-(1)/(y^(2))=(4)/(a^(2))` |
|
| 45. |
Locus of perpendicular from center upon normal to the hyperbola `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1` isA. `(x^(2)-+y^(2))^(2)((a^(2))/(x^(2))+(b^(2))/(y^(2))) =(a^(2)-b^(2))^(2)`B. `(x^(2)+y^(2))^(2)((a^(2))/(x^(2))-(b^(2))/(y^(2)))=(a^(2)+b^(2))^(2)`C. `(x^(2)+y^(2))^(2) ((x^(2))/(a^(2))-(y^(2))/(b^(2))) =(a^(2)+b^(2))^(2)`D. None of these |
|
Answer» Correct Answer - B Let the foot of perpendicular from center upon any normal be `P(h,k)` Then equation of normal to hyperbola is `(y-k) =- (h)/(k) (x-h)` or `hx + ky = h^(2) +k^(2)` (1) Also normal at any point `R(a sec theta, tan theta)` on the hyperbola is `(ax)/(sec theta) +(by)/(tan theta) = a^(2) + b^(2)` (2) Comparing ratio coefficients of equations (1) and (2), We get `((a)/(sec theta))/(h)=((b)/(tan theta))/(k) =(a^(2)+b^(2))/(h^(2)+k^(2))` `rArr sec theta = (a(h^(2)+k^(2)))/(h(a^(2)+b^(2)))` and `tan theta = (b(h^(2)+k^(2)))/(k(a^(2)+b^(2)))` Squaring and subtracting `(x^(2)+y^(2))^(2) ((a^(2))/(x^(2))-(b^(2))/(y^(2))) = (a^(2)+b^(2))^(2)` |
|
| 46. |
Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola `16y^2 -9 x^2 = 1` isA. `x^(2)+y^(2)=9`B. `x^(2)+y^(2)=1//9`C. `x^(2)+y^(2)=7//144`D. `x^(2)+y^(2)=1//16` |
|
Answer» Correct Answer - D `(y^(2))/(1//16)-(x^(2))/(1//9)=1` Locus will be the auxiliary circle `x^(2)+y^(2)=(1)/(16)` |
|
| 47. |
If `a`hyperbola passes through the foci of the ellipse `(x^2)/(25)+(y^2)/(16)=1`. Its transverse and conjugate axes coincide respectively with themajor and minor axes of the ellipse and if the product of eccentricities ofhyperbola and ellipse is 1 thenthe equation ofhyperbola is `(x^2)/9-(y^2)/(16)=1`b. the equation of hyperbola is `(x^2)/9-(y^2)/(25)=1`c. focus of hyperbola is (5, 0)d. focusof hyperbola is `(5sqrt(3),0)`A. the equation of hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1`B. the equation of the hyperbola is `(x^(2))/(9)-(y^(2))/(25)=1`C. the vertex of the hyperbola is (5, 0)D. the vertex of the hyperbola is `(5sqrt3, 0)` |
|
Answer» Correct Answer - A::C for given ellipse `(x^(2))/(25)+(y^(2))/(16)=1` we have `e=sqrt(1-(16)/(25))=(3)/(5)` Hence, the eccentricity of the hyperbola is `5//3`. Let the hyperbola be `(x^(2))/(A^(2))-(y^(2))/(B^(2))=1` Then `B^(2)=A^(2)(e^(2)-1)=A^(2)((25)/(9)-1)=(16)/(9)A^(2)` Therefore, the equation of the hyperbola is `(x^(2))/(A^(2))-(9y^(2))/(16A^(2))=1` As it passes through (3, 0), we get `A^(2)=9 and B^(2)=16.` The equation is `(X^(2))/(9)-(y^(2))/(16)=1` The foci of the hyperbola are `(pmae, 0)-=(pm 5, 0).` The vertex of the hyperbola is (3, 0). |
|
| 48. |
If the foci of a hyperbola lie on `y=x`and one of the asymptotes is `y=2x ,`then the equation of the hyperbola, given that it passes through (3,4), is`x^2-y^2-5/2x y+5=0``2x^2-2y^2+5x y+5=0``2x^2+2y^2+5x y+10=0`none of these |
|
Answer» Correct Answer - `2x^(2)+2y^(2)-5xy+10=0` The foci of the hyperbola lie on y = x. So, the major axis is y = x. The major axis of hyperbola bisects the asymptote. Therefore, the equation of other asymptote is x = 2y. Thus, the equation of hyperbola is `(y-2x)(x-2y)+k=0.` Given that it passes through (3,4), we get `k=-10.` Hence, the required equation is `2x^(2)+2y^(2)-5xy+10=0` |
|
| 49. |
If the sum of the slopes of the normal from a point `P`to the hyperbola `x y=c^2`is equal to `lambda(lambda in R^+)`, then the locus of point `P`is`x^2=lambdac^2`(b) `y^2=lambdac^2``x y=lambdac^2`(d) none of these |
|
Answer» Correct Answer - `x^(2)=lambdac^(2)` Equation of hyperbola is `xy = c^(2)`. Differentiating w.r.t. x, we get `(dy)/(dx)=-(y)/(x)` Thus, slope of normal at any point on the hyperbola is `-(dx)/(dy)=(x)/(y)`. If point on the hyperbola is (x, y)`-=` (ct, c/t), then `-(dx)/(dy)=t^(2)`. So, equation of normal at point (ct, c/t) is `y-(c)/(t)=t^(2)(x-ct)` `rArr" "ct^(4)=xt^(3)+yt-c=0` If this normal passes through the point P(h, k) on the plane, then we have `ct^(4)-ht^(3)+kt-c=0" (1)"` This is polynomial equation of four degree in variable t. So, we can get maximum four real roots of the equation. Hence, maximum four normals can be drawn from point `p(h,k)`. Now, given that sum of the slopes of normals is `lambda`. i.e., `t_(1)^(2)+t_(2)^(2)+t_(3)^(2)+t_(4)^(2)=lambda`, where `t_(1),t_(2),t_(3) and t_(4)` are roots of equation (1). `therefore" "(t_(1)+t_(2)+t_(3)+t_(4))^(2)-sumt_(1)t_(2)=0` `rArr" "((h)/(c))^(2)-0=lambda` [From equation (1), using sum of roots and sum of product of roots taking two at a time] `rArr" "h^(2)=c^(2)lambda` Therefore, required locus is `x^(2)=lambdac^(2).` |
|
| 50. |
A hyperbola passes through the point `P(sqrt(2),sqrt(3))`and has foci at `(+-2,0)dot`Then the tangent to this hyperbola at `P`also passes through the point : |
|
Answer» `x^2/a^2-y^2/b^2=1` `x^2/a^2-y^2/(a^2(e^2-1))=1` `x^2/a^2=y^2/(4-a^2)=1` `(sqrt2)^2/a^2-(sqrt3)^2/(4-a^2)=1` `2/a^2-3/(4-a^2)=1` `(8-2a^2-3a^2)/(a^2(4-a^2))=1` `8-2a^2-3a^2-4a^2+a^4=0` Let`a^2=t` `t^2-9t+8=0` `(t-1)(t-8)=0` `t=1,8` `a^2=1` `|a|=1` `e=2` `b^2=1(4-1)=3` `x^2/1-y^22/3=1` `sqrt2/1x-sqrt3/3y=1` `sqrt2x-y/sqrt3=1` `4-3=1` `1=1` Option 4 is correct. |
|