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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.
| 101. |
A tangent to the hyperbola `x^(2)-2y^(2)=4` meets x-axis at P and y-aixs at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is origin).Find the locus of R. |
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Answer» Equation of tangent at point `(a sec theta, b tan theta) ` is `(x)/(a) sec theta-(y)/(b) tan theta=1` It meets axis at `P(a cos theta, 0 ) and Q(0, -b cot theta).` Now, rectangle OPRQ is completed. Let the coordinates of point R be (h, k). `therefore" "h=a cos theta and k=-bcot theta` `therefore" "sectheta=(a)/(h) and tan theta=(-b)/(k)` Squaring and subtracting, we get `(a^(2))/(h^(2))-(b^(2))/(k^(2))=1.` So, required locus is `(4)/(x^(2))-(2)/(y^(2))=1`. |
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| 102. |
The eccentricity of the hyperbola `9x^(2)-16y^(2)+72x-32y-16=0`, isA. `(5)/(4)`B. `(4)/(5)`C. `(9)/(16)`D. `(16)/(9)` |
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Answer» We have, `9(x^(2)+8x)-16(y^(2)-2y)=16` `implies9(x+4)^(2)-16(y+1)^(2)=144` `implies((x+4)^(2))/(16)-((y+1)^(2))/(9)=1` Here, `a^(2)=16` and `b^(2)=9`. `:.e^(2)=1+(b^(2))/(a^(2))=1+(9)/(16)=(25)/(16)impliese=(5)/(4)` |
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| 103. |
Consider the hyperbola `(X^(2))/(9)-(y^(2))/(a^(2))=1` and the circle `x^(2)+(y-3)=9`. Also, the given hyperbola and the ellipse `(x^(2))/(41)+(y^(2))/(16)=1` are orthogonal to each other. The number of points on the hyperbola and the circle from which tangents drawn to the circle and the hyperbola, respectively, are perpendicular to each other is |
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Answer» Correct Answer - C Ellipse and hyperbola are orthogonal and therefore, they are confocal. `"So, "a_(h)e_(h)=a_(e)e_(e)` `rArr" "a^(2)+9=41-16` `rArr" "9+a^(2)=25` `rArr" "a^(2)=16` Thus, hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1.` So, common tangents to the circle and hyperbola are `x = pm 3`. Director circle of hyperbola does not exist as `a lt b.` Director circle of circle is `x^(2)+(y-3)^(2)=18` `rArr" "x^(2)+y^(2)-6y-9=0` This meets the hyperbola `16x^(2)-9y^(2)=144` at four points from where tangents drawn to the circle `x^(2)+(y-3)^(2)=9` are perpendicular to each other. Let midpoint of AB be (h,k). So, equation of line AB is `hx+ky-3(y+k)=h^(2)+k^(2)-6k`. Since tangents at C and D intersect at the directrix, CD is the focal chord of hyperbola. So, AB passes through focus of the hyperbola and that is `(pm5,0)`. Therefore, required lacus is `x^(2)+y^(2)pm5x-3y=0`. |
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| 104. |
An ellipse intersects the hyperbola `2x^2-2y =1` orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (b) the foci of ellipse are `(+-1, 0)` (a) equation of ellipse is `x^2+ 2y^2 =2` (d) the foci of ellipse are `(t 2, 0)` (c) equation of ellipse is `(x^2 2y)`A. the equation of the ellipse is `x^(2)+2y^(2)=1`B. the foci of the ellipse are `(pm1,0)`C. the equation of the ellipse is `x^(2)+2y^(2)=4`D. the foci of the ellipse are `(pmsqrt2,0)` |
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Answer» Correct Answer - A::B Let the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`. Since ellipse intersects hyperbola orthogonally. The ellipse and hyperbola will be confocal. So, comparing coordinates of foci `(pma+(1)/(sqrt2),0)=(pm1,0)` `"or "a=sqrt2 and e=(1)/(sqrt2)` Now, `b^(2)=a^(2)(1-e^(2))` `"or "b^(2)=1` Therefore, the equation of the ellipse is `(x^(2))/(2)+(y^(2))/(1)=1` |
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| 105. |
Consider the hyperbola `(X^(2))/(9)-(y^(2))/(a^(2))=1` and the circle `x^(2)+(y-3)=9`. Also, the given hyperbola and the ellipse `(x^(2))/(41)+(y^(2))/(16)=1` are orthogonal to each other. Combined equation of pair of common tangents between the hyperbola and the circle is given beA. `x^(2)-y^(2)=0`B. `x^(2)-9=0`C. `9y^(2)-19x^(2)=0`D. No common tangent. |
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Answer» Correct Answer - B Ellipse and hyperbola are orthogonal and therefore, they are confocal. `"So, "a_(h)e_(h)=a_(e)e_(e)` `rArr" "a^(2)+9=41-16` `rArr" "9+a^(2)=25` `rArr" "a^(2)=16` Thus, hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1.` So, common tangents to the circle and hyperbola are `x = pm 3`. Director circle of hyperbola does not exist as `a lt b.` Director circle of circle is `x^(2)+(y-3)^(2)=18` `rArr" "x^(2)+y^(2)-6y-9=0` This meets the hyperbola `16x^(2)-9y^(2)=144` at four points from where tangents drawn to the circle `x^(2)+(y-3)^(2)=9` are perpendicular to each other. Let midpoint of AB be (h,k). So, equation of line AB is `hx+ky-3(y+k)=h^(2)+k^(2)-6k`. Since tangents at C and D intersect at the directrix, CD is the focal chord of hyperbola. So, AB passes through focus of the hyperbola and that is `(pm5,0)`. Therefore, required lacus is `x^(2)+y^(2)pm5x-3y=0`. |
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| 106. |
Find the equation of the hyperbola whose eccentricity is `sqrt(2)`, focus is (a,0) and directrix is the line x = `(a)/(2)`. |
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Answer» Correct Answer - `2x^(2)-2y^(2)= a^(2)` |
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| 107. |
Find the equation of the hyperbola whose eccentricity is `sqrt(2)`, focus is (3,1) and equation of directrix is 2x + y - 1 = 0. |
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Answer» Correct Answer - `3x^(2)-3y^(2) +8xy + 22x + 6y - 48 =0` |
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| 108. |
Find the equation of the hyperbola whose eccentricity is `sqrt(3)`, focus is at (2,3) and equation of directrix is x + 2y = 1. |
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Answer» Correct Answer - `2x^(2)-9y^(2) - 12xy - 14x - 18y + 62 =0` |
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| 109. |
Equation of the rectangular hyperbola whose focus is `(1,-1)` and the corresponding directrix is `x-y+1=0`A. `x^(2)-y^(2)=1`B. `xy=1`C. `2xy-4x+4y+1=0`D. `2xy+4x-4y-1=0` |
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Answer» Correct Answer - C Eccentricity of rectangular hyperbola is `sqrt2`. So, equation of hyperbola is `sqrt((x-1)^(2)+(y+1)^(2))=sqrt2(|x-y+1|)/(sqrt(1^(2)+(-1)^(2)))` `rArr" "(x-1)^(2)+(y+1)^(2)=(x-y+1)^(2)` `rArr" "2xy-4x+4y+1=0` |
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| 110. |
Circles are drawn on chords of the rectangular hyperbola xy = 4 parallel to the line y = x as diameters. All such circles pass through two fixed points whose coordinates areA. (2, 2)B. (2, -2)C. `(-2, 2)`D. `(-2, -2)` |
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Answer» Correct Answer - A::D The circle with points `P(2t_(1),2//t_(1))` and `Q(2t_(2),2//t_(2))` as diameter is given by `(x-2t_(1))(x-2t_(2))+(y-(2)/(t_(1)))(y-(2)/(t_(2)))=1" (1)"` Also, the slope of PQ is given by `-(1)/(t_(1)t_(2))=1 or t_(1)t_(2)=-1` Hence, from (1), the circle is `(x^(2)+y^(2)-8)-2(t_(1)+t_(2))(x-y)=0` which is of the form `S+lambdaL=0`. Hence, circles pass through the points of intersection of the circle `x^(2)+y^(2)-8=0` and the line x = y. The point of intersection are (2, 2) and `(-2, -2)`. |
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| 111. |
The locus of the centre of a variable circle touching two circles of radii `r_(1)`, `r_(2)` externally , which also touch eath other externally, is conic. If `(r_(1))/(r_(2))=3+2sqrt(2)`, then eccentricity of the conic, isA. `1`B. `sqrt(2)`C. `1//2`D. `2sqrt(2)` |
| Answer» Correct Answer - option 2 | |
| 112. |
The distance of the focus of `x^(2)-y^(2) =4`, from the directrix, which is nearer to it, isA. `2sqrt(2)`B. `sqrt(2)`C. `4sqrt(2)`D. `8sqrt(2)` |
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Answer» Correct Answer - B `x^(2) -y^(2) =4` is rectangular hyperbola. `:. e = sqrt(2)` Required distance `= ae - a//e = 2 sqrt(2) -2//sqrt(2) = sqrt(2)` |
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| 113. |
Show that the midpoints of focal chords of a hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`lie on another similar hyperbola. |
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Answer» Let the chord AB be bisected at point `P(h,k).` So, equation of chord AB is `(hx)/(a^(2))-(ky)/(b^(2))=(h^(2))/(a^(2))-(k^(2))/(b^(2))" (Using T = S"_(1)")"` Let it pass through the focus (ae, 0). (1) `therefore" "(he)/(a)=(h^(2))/(a^(2))-(k^(2))/(b^(2))` Therefore, locus of point P is (2) `(x^(2))/(a^(2))-(y^(2))/(b^(2))=(ex)/(a)` `rArr" "(1)/(a^(2))[x^(2)-aex]-(y^(2))/(b^(2))=0` `rArr" "((x-(ae)/(2))^(2))/(a^(2))-(y^(2))/(b^(2))=(e^(2))/(4)` This is also hyperbola with eccentricity e. |
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| 114. |
Find the equation of the hyperbola whose conjugateaxis is 5 and the distance between the foci is 13. |
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Answer» Length of congugate axis of hyperbola ` = 5` `=>2b = 5 =>b = 5/4` `=>b^2 = 25/4->()` Distance berween the foci ` =13` `=>2ae = 13` Squaring both sides, `=>4a^2e^2 = 13^2` `=>4a^2(1+b^2/a^2) = 169` `=>a^2+b^2 = 169/4` `=>a^2+25/4 = 169/4` `=>a^2 = 169/4 -25/4 = 144/4` `=>a^2 = 36` Equation of a hyperbola is given by, `x^2/a^2-y^2/b^2 = 1` `:.` Equation of the given hyperbola, `=>x^2/36-(4y^2)/25 = 1` |
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| 115. |
A point moves on a plane in such a manner that the difference of its distances from the points (4,0) and (-4,0) is always constant and equal to `4sqrt(2)` show that the locus of the moving point is a rectangular hyperbola whose equation you are to determine. |
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Answer» Correct Answer - `x^(2)-y^(2) = 8` |
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| 116. |
In X-Y plane, the path defined by the equation `(1)/(x^(m))+(1)/(y^(m)) +(k)/((x+y)^(n)) =0`, isA. a parabola if `m = (1)/(2), k =- 1, n =0`B. a hyperbola if `m =1, k =- 1, n=0`C. a pair of lines if `m = k = n =1`D. a pair of lines if `m = k =- 1, n =1` |
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Answer» Correct Answer - A::B::C::D (a) `sqrt(x) + sqrt(y) =1` `rArr + y+2 sqrt(xy) =1` `rArr 4xy = (1-x-y)^(2)` (b) `(1)/(x) +(1)/(y) =1 rArr xy - x - y=0` is a hyperbola (c ) `(1)/(x) +(1)/(y) + (1)/(x+y) =0` `rArr x^(2) + 3xy + y^(2) =0`, which is a pair of lines. (d) `x + y -(1)/(x+y) =0` `rArr (x+y)^(2) =1` `rArr x+y = +-1` which is a pair of lines. |
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| 117. |
If `(x^(2))/(36)-(y^(2))/(k^(2))=1` is a hyperbola, then which of the following points lie on hyperbola?A. `(3,1)`B. `(-3,1)`C. `(5,2)`D. `(10,4)` |
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Answer» Correct Answer - D For hyperbola `(x^(2))/(36) -(y^(2))/(k^(2)) =1, a = 6` `:.` Abscissa of any point on hyperbola must be `ge 6` `:. (10,4)` can lie on hyperbola |
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| 118. |
If the distance between the foci of a hyperbola is`16`and its eccentricity is `sqrt(2)`, then obtain its equation. |
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Answer» Distance between the foci of the hyperbola, `2ae =16` Eccentricity of the hyerbola, `e = sqrt2` `:. (2ae)/e = 16/sqrt2` `a = 8/sqrt2 = 4sqrt2` Now, `e^2 = 1+ b^2/a^2` `=>sqrt2^2 = 1 + b^2/(4sqrt2)^2` `=>2 = 1+b^2/32` `=>b^2 = 32(2-1)` `=>b^2 = 32` Equation of a hyperbola is given by, `x^2/a^2-y^2/b^2 = 1` `:.` Equation of the given hyperbola, `=>x^2/32-y^2/32 = 1` `=>x^2-y^2 = 32.` |
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| 119. |
The ellipse `(x^(2))/(25)+(y^(2))/(16)=1` and the hyperbola `(x^(2))/(25)-(y^(2))/(16) =1` have in commonA. centre and vertices onlyB. centre, foci and verticesC. centre, foci and directricesD. centre only |
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Answer» Correct Answer - A For ellipse `(x^(2))/(25) + (y^(2))/(16) =1, a = 5` and `b = 4` For hyperbola `(x^(2))/(25) - (y^(2))/(16) =1, a = 5` and `b = 4` For ellipse `a^(2)e^(2) = a^(2)-b^(2) = 25 - 16 =9` For hyperbola `a^(2)e^(2) = a^(2)+b^(2) = 25 + 16 = 41` `:.` center and vertices only are common. |
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| 120. |
Find the equation of tangents to hyperbola `x^(2)-y^(2)-4x-2y=0` having slope 2. |
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Answer» We have hyperbola `x^(2)-y^(2)-4x-2y=0` `"or "(x-2)^(2)-(y+1)^(2)=3` `"or "((x-2)^(2))/(3)-((y+1)^(2))/(3)=1" (1)"` Equation of tangents of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` having slope m is given by `y=mxpm sqrt(a^(2)m^(2)-b^(2))` So, equation of tangents to hyperbola (1) having slope 2 is given by `y+1=2(x-2)pmsqrt(3xx4-3)` `"or "y+1=2x-4pm3` Therefore, the equations of tangents to given hyperbola are `2x-y-2=0` and `2x-y-8=0.` |
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| 121. |
The tangents and normal at a point on `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =1` cut the y-axis A and B. Then the circle on AB as diameter passes throughA. one of the vertex of the hyperbolaB. one of the foot of directrix on x-axis of the hyperbolaC. foci of the hyperbolaD. none of these |
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Answer» Correct Answer - C Equation of tangent at point `P(theta)` is `(sec theta)/(a) x -(tan theta)/(b) y =1` `:. A (0,-b cot theta)` Equation of normal at point `P(theta)` is `a cos theta x + b cot theta y = a^(2) +b^(2)` `:. B(0,(a^(2)+b^(2))/(b cot theta))` Equation of circle as AB as a diameter is `(x-0) (x-0) + (y+b cot theta) (y-(a^(2)+b^(2))/(b cot theta)) =0` or `x^(2)+ (y+b cot theta) (y-(a^(2)e^(2))/(b cot theta)) =0` Clearly this passes through foci (ae,0) |
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| 122. |
The equation of a hyperbola with co-ordinate axes as principal axes, and the distances of one of its vertices from the foci are 3 and 1 can beA. `3x^(2) -y^(2) =3`B. `x^(2)-3y^(2) +3 =0`C. `x^(2)-3y^(2) -3 =0`D. none of these |
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Answer» Correct Answer - A::B Consider `(x^(2))/(a^(2)) - (y^(2))/(b^(2)) =1` Let one of the vertices be `(a,0)` Foci are `(+- ae,0)` According to the question we have `ae - a = 1` and `ae + a=3` Solving we get `a = 1` and `e =2` `:. e^(2) =1 + (b^(2))/(a^(2)) rArr b^(2) =3` `:.` Equation of hyperbola is `(x^(2))/(1) -(y^(2))/(3) =1` or `3x^(2) - y^(2) =3` For `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =-1` is `e^(2) =1 + (a^(2))/(b^(2)) rArr b^(2) = (1)/(3)` `:.` Hyperbola will be `x^(2) - 3y^(2) + 3=0` |
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| 123. |
Find the equation of the hyperbola with eccentricity `sqrt(2)` and the distance between whose foci is 16. |
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Answer» Correct Answer - `x^(2)-y^(2)=32` `2c=16hArr c=8 ." " Also , e=sqrt(2)` `"Now" , e=(c )/(a)implies a=(c )/(e) =(8)/(sqrt(2))=4sqrt(2)`. `therefore b^(2)=(c^(2)-a^(2))=(64-32)=32.` Thus , `a^(2)=32 and b^(2)=32.` |
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| 124. |
If the normal at `P(asectheta,btantheta)` to the hyperbola `x^2/a^2-y^2/b^2=1` meets the transverse axis in G then minimum length of PG isA. `(b^(2))/(a)`B. `|(a)/(b)(a+b)|`C. `|(a)/(b)(a-b)|`D. `|(a)/(b)(a-b)|` |
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Answer» Correct Answer - A Equation of the normal is `ax cos theta + by cot theta = a^(2) +b^(2)` The normal at P meets the coordinate axes at `G((a^(2)+b^(2))/(a)sec theta,0)` and `g (0,(a^(2)+b^(2))/(b)tan theta)` `:. PG^(2) = ((a^(2)+b^(2))/(a)sec theta -a sec theta)^(2)+(b tan theta -0)^(2)` `PG^(2) =(b^(2))/(a^(2)) (b^(2) sec^(2) theta + a^(2) tan^(2) theta)` When `tan theta =0` `PG =(b^(2))/(a)` |
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| 125. |
The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8 isA. `(x^(2))/(12)-(y^(2))/(4)=1`B. `(x^(2))/(4)-(y^(2))/(12)=1`C. `(x^(2))/(8)-(y^(2))/(2)=1`D. `(x^(2))/(16)-(y^(2))/(9)=1` |
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Answer» Correct Answer - B Distance between foci = 8 `:. 2ae = 8` also `e = 2, :. 2a = 4` `rArr a = 2 rArr a^(2) = 4 :. b^(2) = 4(4-1) = 12` `:.` Equation of hyperbola is `(x^(2))/(4) -(y^(2))/(12) =1`. |
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| 126. |
Find the coordinates of the foci of the hyperboala `x^(2) - y^(2) + 1 =0` |
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Answer» Correct Answer - `(0,pmsqrt(2))` |
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| 127. |
Find the equation of tangent to the hyperbola `y=(x+9)/(x+5) ` which passes through `(0, 0)` originA. `x+25y=0`B. `x+y=0`C. `5x-y=0`D. `x-25y=0` |
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Answer» Correct Answer - A::B `y=(x+9)/(x+5)=1+(4)/(x+5)` `therefore" "(dy)/(dx)=(-4)/((x+5)^(2))` At `(x_(1),y_(1)),(dy)/(dx)=(-4)/((x_(1)+5)^(2))` Equation of tangent at `(x_(1),y_(1))`, `y-y_(1)=(-4)/((x_(1)+5)^(2))(x-x_(1))` `rArr" "y-1-(4)/(x_(1)+5)=(-4)/((x_(1)+5)^(2))(x-x_(1))` Since it passes through (0, 0), `(x_(1)+5)^(2)+4(x_(1)+5)+4x_(1)=0` `rArr" "x_(1)^(2)+18x_(1)+45=0` `therefore" "x_(1)=-15or x_(1)=-3` So, equations of required tangents are `x+25y=0 or x+y=0`. |
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| 128. |
The equation of a tangent to the hyperbola `3x^(2)-y^(2)=3`, parallel to the line `y = 2x +4` isA. `y = 2x +3`B. `y = 2x+1`C. `y = 2x+4`D. `y = 2x+2` |
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Answer» Correct Answer - B `3x^(2) - y^(2) = 3, (x^(2))/(1) -(y^(2))/(3) =1` Equation of tangent in terms of slope. `y = mx +- sqrt((m^(2)-3))` Here, `m =2`, then `y = 2x +-1` |
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| 129. |
The absolute value of slope of common tangents to parabola `y^(2) = 8x` and hyperbola `3x^(2) -y^(2) =3` isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B Tangent to `y^(2)=8x` is `y = mx + (2)/(m)` Tangent to `(x^(2))/(1) -(y^(2))/(3) =1` is `y = mx+-sqrt(m^(2)-3)` on comparing, we get `m = +-2` |
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| 130. |
The equation `(x-alpha)^2+(y-beta)^2=k(lx+my+n)^2` representsA. a parabola for `k lt(l^(2)+m^(2))^(-1)`B. an ellipse for `0 lt k lt(l^(2)+m^(2))^(-1)`C. a hyperbola for `k gt (l^(2)+m^(2))^(-1)`D. a point circle for k = 0 |
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Answer» Correct Answer - B::C::D `(x-alpha)^(2)+(gamma-beta)^(2)=k(lx+my+n)^(2)` `"or "sqrt((x-alpha)^(2)+(y-beta)^(2))=sqrtksqrt(l^(2)+m^(2))((lx+my+n))/(sqrt(l^(2)+m^(2)))` `"or "(PS)/(PM)=sqrtksqrt(l^(2)+m^(2))` where P(x, y) is any point on the curve. Fixed `S(alpha, beta)` is focus and fixed line `lx+my+n=0` is directrix. If `k(l^(2)+m^(2))=1, P` lies on a parabola. If `k(l^(2)+m^(2))lt1`, P lies on an ellipse. If `k(l^(2)+m^(2))gt1`, P lies on a hyperbola. If k = 0, P lies on a point circle. |
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| 131. |
Find the position of the point (7,2) with respect to the hyperbola `9x^(2) - 16y^(2) = 144`. |
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Answer» Correct Answer - The point (7,2) lies inside the given hyperbola |
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| 132. |
If e is the eccentricity of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` and `theta` is the angle between the asymptotes, then `cos.(theta)/(2)` is equal toA. `(1-e)/(e)`B. `(2)/(e ) -e`C. `(1)/(e )`D. `(2)/(e )` |
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Answer» Correct Answer - C `theta = 2 tan^(-1).(b)/(a) rArr tan.(theta)/(2) = (b)/(a)` `rArr cos.(theta)/(2) = (a)/(sqrt(a^(2)+b^(2))) = (1)/(sqrt(1+(b^(2))/(a^(2)))) = (1)/(e)` |
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| 133. |
Let `P(a sectheta, btantheta) and Q(aseccphi , btanphi)` (where `theta+phi=pi/2` be two points on the hyperbola `x^2/a^2-y^2/b^2=1` If `(h, k)` is the point of intersection of the normals at `P and Q` then `k` is equal to (A) `(a^2+b^2)/a` (B) `-((a^2+b^2)/a)` (C) `(a^2+b^2)/b` (D) `-((a^2+b^2)/b)`A. `(a^(2)+b^(2))/(a)`B. `-((a^(2)+b^(2))/(a))`C. `(a^(2)+b^(2))/(b)`D. `-((a^(2)+b^(2))/(b))` |
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Answer» Firstly, we obtain the slope fo normal to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 " at " (a sec theta , b tan theta)`. On differentiating w.r.t.x, we get `(2x)/(a^(2))-(2y)/(b^(2))xx(dy)/(dx)=0 rArr(dy)/(dx)=(b^(2))/(a^(2))(x)/(y)` Slope for normal at the point `(a sec theta , b tan theta )` will be `-(a^(2)b tan theta )/(b^(2)a sec theta)= -(a)/(b) sin theta ` ` therefore ` Equation of normal at `(a sec theta, b tan theta )` is `y-b tan theta= -(a)/(b) sin theta (x-a sec theta)` `rArr (a sin theta )x + "by"=(a^(2)+b^(2))tan theta` `rArr ax+b "cosec" theta =(a^(2)+b^(2)) sec theta " ...(i)" ` Similarly, equation of normal to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` at `(a sec phi, b tan phi) " is " ax+b y " cosec "phi=(a^(2)+b^(2))sec phi " ... (ii)" ` On subtracting Eq. (ii) from Eq. (i), we get `b("cosec" theta -"cosec" phi)y=(a^(2)+b^(2))(sec theta-sec phi)` `rArr y=(a^(2)+b^(2))/(b)*(sec theta-sec phi)/("cosec"theta-"cosec"phi)` But `(sec theta-sec phi)/("cosec"theta-"cosec"phi)=(sec theta-sec(pi//2-theta))/("cosec"theta-"cosec"(pi//2-theta)) " " [ because phi+theta=pi//2]` `=(sec theta -"cosec"theta)/(sec theta-sec theta)= -1` Thus, `y= -((a^(2)+b^(2))/(b)),i.e.k=-((a^(2)+b^(2))/(b))` |
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| 134. |
If the centre, vertex and focus of a hyperbola be (0,0), (4,0) and (6,0) respectively, then the equation of the hyperbola isA. `4x^(2) -5y^(2) = 8`B. `4x^(2)-5y^(2) = 80`C. `5x^(2)-4y^(2) =80`D. `5x^(2)-4y^(2) =8` |
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Answer» Correct Answer - C Centre (0,0), vertex (4,0) `rArr a = 4` and focus (6,0) `rArr ae = 6 rArr e = (3)/(2)`. Therefore `b = sqrt(20)` Hence required equation is `(x^(2))/(16) -(y^(2))/(20) =1` i.e., `5x^(2) - 4y^(2) = 80` |
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| 135. |
If P is any point common to the hyperbola `(x^(2))/(16)-(y^(2))/(25)=1` and the circle having line segment joining its foci as diameter then sum of focal distances of point P isA. `6sqrt(2)`B. `2sqrt(66)`C. 16D. 8 |
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Answer» Correct Answer - B Foci of hyperbola are `(+- sqrt(41),0)` `:. P` lies on the circle `x^(2) + y^(2) = 41` Any point on hyperbola is `(4 sec theta, 5 tan theta)` `rArr 16 sec^(2) theta + 25 tan^(2) theta = 41` `rArr tan theta = (5)/(sqrt(41))` and `sec theta = sqrt((66)/(41))` `:. PF_(1) + PF_(2) = e (4 sec theta -(a)/(e)) +e (4 sec theta +(a)/(e))` `= 8 esec theta = 2 sqrt(66)` |
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| 136. |
The distance between the two foci of the hyperbola `x = 6 sec phi , y = 6 tan phi` is -A. `16sqrt(20`B. `12sqrt(2)`C. `8sqrt(2)`D. none of these |
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Answer» Correct Answer - B |
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| 137. |
A tangent to the hyperbola `y = (x+9)/(x+5)` passing through the origin isA. `x+ 25y =0`B. `5x+y =0`C. `5x-y =0`D. `x -25y =0` |
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Answer» Correct Answer - A `y = (x+9)/(x+5) =1+ (4)/(x+5)` `(dy)/(dx)` at `(x_(1),y_(1)) = (-4)/((x_(1)+5)^(2))` `:.` Equation of tangent `y - y_(1) = (-4)/((x_(1)+5)^(2)) (x-x_(1))` `y -1 - (4)/(x_(1)+5) = (-4)/((x_(1)+5)^(2)) (x-x_(1))` Since it passes through (0,0) `(x_(1)+5)^(2) + 4(x_(1)+5) + 4x_(1) =0` `x_(1) =- 15` or `x_(1) =-3`. So equations are `x + 25y = 0` or `x + y =0`. |
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| 138. |
If `(5,12)a n d(24 ,7)`are the foci of a hyperbola passing through the origin, then`e=(sqrt(386))/(12)`(b) `e=(sqrt(386))/(13)``L R=(121)/6`(d) `L R=(121)/3`A. `e=sqrt(386)/12`B. `e=sqrt(386)/13`C. `LR=121//6`D. `LR=121//3` |
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Answer» Correct Answer - A::C We have foci `S_(1)(5, 12) and S_(2)(24, 7)`. Hyperbola passes through P(0, 0). `therefore" "|PS_(1)-PS_(2)|=2a` `2a=K` `"or "2a=|sqrt((24-0)^(2)+(7-0)^(2))-sqrt(12^(2)+5^(2))|=12` `therefore" "a=6` `S_(1)S_(2)=2ae=sqrt((24-5)^(2)+(12-7)^(2))=sqrt(386)` `therefore" "e=(sqrt(386))/(12)` `LR = (2b^(2))/(a)=(2a^(2)(e^(2)-1))/(a)` `=2xx6((386)/(144)-1)=(121)/(6)` |
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| 139. |
Find the length of the transverse and conjugate axes of the hyperbola `9x^(2) - 16y^(2) = 144`. Write down the equation of the hyperbola conjugate to it and find the eccentricities of both the hyperbolas. |
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Answer» Correct Answer - 8 and 6 and `e = 5/4, 16y^(2)-9x^(2) = 144,e = 5/3` |
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| 140. |
The slopes of the common tangents of the hyperbolas `(x^(2))/(9)-(y^(2))/(16)=1` and `(y^(2))/(9)-(x^(2))/(16)=1`, areA. `+-2`B. `+-1`C. `+-1//2`D. none of these |
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Answer» Given hyperbolas are : `(x^(2))/(9)-(y^(2))/(16)=1`……..`(i)` and `(x^(2))/(16)-(y^(2))/(9)=-1`……`(ii)` The equation of any tangent to `(i)` is `y=mx+-sqrt(9m^(2)-16)` If it touches `(ii)`, then `9m^(2)-16=9-16m^(2)`[If `y=mx+c` touches `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1` then `c^(2)=b^(2)-a^(2)m^(2)`] `impliesm=+-1` |
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| 141. |
The number of points from where a pair of perpendiculartangents can be drawn to the hyperbola, `x^2 sec^2 alpha -y^2 cosec^2 alpha=1, alpha in (0, pi/4)`, is (A) 0 (B) 1 (C) 2 (D) infinite |
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Answer» Correct Answer - D `(x^(2))/(cos^(2)alpha) - (y^(2))/(sin^(2)alpha) =1` Locus of perpendicular tangents is director circle, `x^(2) + y^(2) = a^(2) - b^(2)` or `x^(2)+y^(2) = cos^(2) alpha - sin^(2) alpha = cos 2 alpha` But `0 lt alpha lt (pi)/(4)` `0 lt 2 alpha lt (pi)/(2)` So there are infinite points. |
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| 142. |
The lines parallel to the normal to the curve `x y=1`is/are`3x+4y+5=0`(b) `3x-4y+5=0``4x+3y+5=0`(d) `3y-4x+5=0`A. `3x+4y+5=0`B. `3x-4y+5=0`C. `4x+3y+5=0`D. `3y-4x+5=0` |
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Answer» Correct Answer - B::D Differentiating xy = 1 w.r.t. x, we have `(dy)/(dx)=-(1)/(x^(2))lt0` Hence, the slope of normal at any point `P(x_(1),y_(1))` is `x_(1)^(2)gt0`. Therefore, the slope of normal must always be positive. Hence, the possible lines are (2) and (4). |
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| 143. |
The equation `(x^(2))/(9-lambda)+(y^(2))/(4-lambda) =1` represents a hyperbola when `a lt lambda lt b` then `(b-a)=`A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C `(9- lambda) (4-lambda) lt 0 rArr 4 lt lambda lt 9` `rArr b -a = 5` |
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| 144. |
nd are inclined at avgicsTangents are drawn from the point `(alpha, beta)` to the hyperbola `3x^2- 2y^2=6` and are inclined atv angle `theta and phi` to the x-axis.If `tan theta.tan phi=2`, prove that `beta^2 = 2alpha^2 - 7`. |
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Answer» Correct Answer - 7 The given hyperbola is `3x^(2)-2y^(2)=6` `"or "(x^(2))/(2)-(y^(2))/(3)=1` Equation of tangent is `y=mx pm sqrt(a^(2)m^(2)-b^(2))` `"or "(y-mx)^(2)=a^(2)m^(2)-b^(2)` Tangents from the point `(alpha, beta)` will be `(beta-malpha)^(2)=2m^(2)-3" "("Since"a^(2)=2 and b^(2)=3)` `"or "m^(2)alpha^(2)+beta^(2)-2malphabeta-2m^(2)+3=0` `m^(2)(alpha^(2)-2)-2alpha betam+beta^(2)+3=0` `m_(1)*m_(2)=(beta^(2)+3)/(alpha^(2)-3)=2=tan thetatan phi` `therefore" "beta^(2)+3=2(alpha^(2)-2)` `"or "2alpha^(2)-beta^(2)=7` |
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| 145. |
If 0 `lt alpha lt (pi)/(2)`, then which of the following is independent of `alpha` in the case of the hyperbola `(x^(2))/(cos^(2)alpha) - (y^(2))/(sin^(2)alpha) = 1`?A. eccentricityB. coordinates of the vertexC. equation of directrixD. abscissa of the foci |
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Answer» Correct Answer - D |
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| 146. |
`(x^(2))/(cos^(2)alpha)-(y^(2))/(sin^(2)alpha)` = `1 (0 lt alpha lt (pi)/(2))` is the equation of a hyperbola, then which of the following(s) is /are depend(s) on `alpha` -A. eccentricityB. abscissa of the focusC. directrixD. vertex |
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Answer» Correct Answer - A,C,D |
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| 147. |
The equation, `2x^2+ 3y^2-8x-18y+35= K` representsA. no locus if k gt 0B. an ellipse if k lt 0C. a point if k = 0D. a hyperbola if k gt0 |
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Answer» Correct Answer - C We have `2x^(2)+3y^(2)-8x-18y+35=k` `"or "2(2x^(2)-4x)+3(y^(2)-6y)+35=k` `"or "2(x-2)^(2)+3(y-3)^(2)=k` For k = 0, we get `2(x-2)^(2)+3(y-3)^(2)=0` which represents the point (2, 3) |
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| 148. |
Find the equation of the hyperbola centre at the origin, transeverse axis on x-axis, passing through the point `(3,sqrt(2),-2)` and having the length of semi-conjugate axis 2. |
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Answer» Correct Answer - `4x^(2)-9y^(2)=36` |
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| 149. |
The differential equation `(dy)/(dx)=(3y)/(2x)`represents a family of hyperbolas (except when it represents a pair oflines) with eccentricity.`sqrt(3/5)`(b) `sqrt(5/3)``sqrt(2/5)`(d) `sqrt(5/2)`A. `sqrt(3//5)`B. `sqrt(5//3)`C. `sqrt(2//5)`D. `sqrt(5//2)` |
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Answer» Correct Answer - B::D (2), (4) `(dx)/(dy)=(3y)/(2x)` `"or "int2xdx=int3ydy` `"or "x^(2)=(3y^(2))/(2)+c` `"or "(x^(2))/(3)-(y^(2))/(2)=(c)/(3)` If c is positive, then `e=sqrt(1+(2)/(3))=sqrt((5)/(3))` If c is negative, then `e=sqrt(1+(3)/(2))=sqrt((5)/(2))` |
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| 150. |
For which of the hyperbolas, can we have more than one pair ofperpendicular tangents?`(x^2)/4-(y^2)/9=1`(b) `(x^2)/4-(y^2)/9=-1``x^2-y^2=4`(d) `x y=44`A. `(x^(2))/(4)-(y^(2))/(9)=1`B. `(x^(2))/(4)-(y^(2))/(9)=-1`C. `x^(2)-y^(2)=4`D. `xy=44` |
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Answer» Correct Answer - B The locus of the point of intersection of perpendicular tangents is director circle for `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` The equation of director circle is `x^(2)+y^(2)=a^(2)-b^(2)`which is real if a gt b. |
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