JEE Mathematics Hyperbola MCQs Set 04

Question. If a hyperbola passes through the focus of the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) and its transverse and conjugate axis coincides with the major and minor axis of the ellipse, and product of their eccentricities is 1, then
(a) equation of hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \)
(b) equation of hyperbola \( \frac{x^2}{9} - \frac{x^2}{25} = 1 \)
(c) focus of hyperbola (5, 0)
(d) focus of hyperbola is \( (5\sqrt{3}, 0) \)
Answer: (a) equation of hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) and (c) focus of hyperbola (5, 0)
Solution:
\( \frac{x^2}{25} + \frac{y^2}{16} = 1 \)
Ecentricity = \( \frac{3}{5} \)
Ecentricity = \( \frac{5}{3} \) and it passes through (± 3, 0)
\( \implies \) its equation \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \)
where \( 1 + \frac{b^2}{9} = \frac{25}{9} \)
\( \implies \) \( b^2 = 16 \)
\( \implies \) \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) and its foci are (± 5, 0)

Comprehension : (3 questions)
Let ABCD be a square of side length 2 units. C_2 is the circle through vertices A, B, C, D and C_1 is the circle touching all the sides of the square ABCD. L is a line through A

Question. If P is a point on C_1 and Q is another point on C_2, then \( \frac{PA^2 + PB^2 + PC^2 + PD^2}{QA^2 + QB^2 + QC^2 + QD^2} \) is equal to
(a) 0.75
(b) 1.25
(c) 1
(d) 0.5
Answer: (a) 0.75
Solution:
Let A, B, C and D be the complex Number \( \sqrt{2} \), \( -\sqrt{2} \), \( \sqrt{2}i \), \( -\sqrt{2}i \) respectively.,
\( \implies \) \( \frac{PA^2 + PB^2 + PC^2 + PD^2}{QA^2 + QB^2 + QC^2 + QD^2} \)
\( = \frac{|z_1 - \sqrt{2}|^2 + |z_1 + \sqrt{2}|^2 + |z_1 - \sqrt{2}i|^2 + |z_1 + \sqrt{2}i|^2}{|z_2 - \sqrt{2}|^2 + |z_2 + \sqrt{2}|^2 + |z_2 - \sqrt{2}i|^2 + |z_2 + \sqrt{2}i|^2} \)
= \( \frac{|z_1|^2 + 2}{|z_2|^2 + 2} = \frac{3}{4} \)

Question. A circle touches the line L and the circle C_1 externally such that both the circles are on the same side of the line, then the locus of centre of the circle is
(a) ellipse
(b) hyperbola
(c) parabola
(d) parts of straight line
Answer: (c) parabola
Solution:
Let c be the centre of the required circle.
Draw a line parallel to L at a distance of \( r_1 \) from it
Now \( CP_1 = AC \)
\( \implies \) C lies oin a parabola

Question. A line M through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts. M at T_2 and T_3 and AC at T_1, then area of \( \Delta T_1T_2T_3 \) is
(a) 1/2 sq. units
(b) 2/3 sq. units
(c) 1 sq. unit
(d) 2 sq. units
Answer: (c) 1 sq. unit
Solution:
Locus of S will be parabola.
\( \because AG = \sqrt{2} \)
\( AT_1 = T_1G = \frac{1}{\sqrt{2}} \)
As A is the focus, \( T_1 \) is the vertex and BD is the direction of parabola]
Also \( T_2T_3 \) is latus rectum
\( \therefore T_2T_3 = 4 \times \frac{1}{\sqrt{2}} \)
Area of \( \Delta T_1T_2T_3 = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{4}{\sqrt{2}} = 1 \)

Question. A hyperbola, having the transverse axis of length \( 2 \sin \theta \), is confocal with the ellipse \( 3x^2 + 4y^2 = 12 \). Then its equation is
(a) \( x^2 \operatorname{cosec}^2\theta - y^2 \sec^2\theta = 1 \)
(b) \( x^2 \sec^2\theta - y^2 \operatorname{cosec}^2\theta=1 \)
(c) \( x^2 \sin^2\theta - y^2 \cos^2\theta = 1 \)
(d) \( x^2 \cos^2\theta - y^2 \sin^2\theta = 1 \)
Answer: (a) \( x^2 \operatorname{cosec}^2\theta - y^2 \sec^2\theta = 1 \)
Solution:
Given \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \)
\( \implies \) \( a = 2 \), \( b = \sqrt{3} \)
\( \implies \) \( e = \frac{1}{2} \)
So that ae = 1
hence the ecentricity \( e_1 \) of hyp. is given by
\( 1 = e_1 \sin \theta \)
\( \implies \) \( e_1 = \operatorname{cosec}\theta \)
\( \implies \) \( b^2 = \sin^2\theta \left( \frac{1 - \sin^2\theta}{\sin^2\theta} \right) = \cos^2\theta \)
Hence the hyp.
\( \frac{x^2}{\sin^2\theta} - \frac{y^2}{\cos^2\theta} = 1 \)
\( \implies \) \( x^2 \operatorname{cosec}^2 \theta - y^2 \sec^2 \theta = 1 \)

Question. Match the statements in Column I with the properties in Column II.
Column – I
(A) Two intersecting circles
(B) Two mutually external circles
(C) Two circles, one strictly inside the other
(D) Two branches of a hyperbola
Column – II
(P) have a common tangent
(Q) have a common normal
(R) do not have a common tangent
(S) do not have a common normal
Answer:
Solution:
(A) When two circle are intersecting they have a common normal and tangent (P)(Q).
(B) Two mutually enternal circles have a common normal and tangent (P)(Q).
(C) When one circle lies inside of other then , they have a common normal but no common tangent. (Q)(R)
(D) Two branches of a hyp. have a common normal but no common tangent. (Q)(R).

Question. Let a and b be non-zero real numbers. Then, the equation \( (ax^2 + by^2 + c) (x^2 - 5xy + 6y^2) = 0 \) represents
(a) four straight lines, when c = 0 and a, b are of the same sign.
(b) two straight lines and a circle, when a = b, and c is of sign opposite to that of a.
(c) two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a
(d) a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a.
Answer: (b) two straight lines and a circle, when a = b, and c is of sign opposite to that of a.
Solution:
\( (ax^2 + by^2 + c) (x^2 - 5xy + 6y^2) = 0 \)
\( ax^2 + by^2 + c = 0 \) or \( x^2 - 5xy + 6y^2 = 0 \)
\( x^2 + y^2 = \left( -\frac{c}{a} \right) \); \( x - 2y = 0 \) and \( x - 3y = 0 \)
If a = b
Hence the given equations represent two straight lines and a circle, when a = b and c is of sign opposite to that of a

Question. Consider a branch of the hyperbola, \( x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0 \) with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is
(a) \( 1 - \sqrt{\frac{2}{3}} \)
(b) \( \sqrt{\frac{3}{2}} - 1 \)
(c) \( 1 + \sqrt{\frac{2}{3}} \)
(d) \( \sqrt{\frac{3}{2}} + 1 \)
Answer: (b) \( \sqrt{\frac{3}{2}} - 1 \)
Solution:
Hyp. is \( \frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1 \)
\( a = 2 \), \( b = \sqrt{2} \)
\( e = \sqrt{\frac{3}{2}} \)
Area = \( \frac{1}{2} a(e - 1) \times \frac{b^2}{a} \)
= \( \frac{1}{2} \times 2 \left( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{2}} \right) \times \frac{2}{2} \)
= \( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{2}} = \sqrt{\frac{3}{2}} - 1 \)

Question. Match the conics in column I with statements/expressions in Column II.
Column-I
(A) Circle
(B) Parabola
(C) Ellipse
(D) Hyperbola
Column-II
(P) The locus of the point (h, k) for which the line hx + ky= 1 touches the circle \( x^2 + y^2 = 4 \)
(Q) Points z in the complex plane satisfying \( |z + 2| - |z - 2|= \pm 3 \)
(R) Points of the conic have parametric representation \( x= \sqrt{3} (1-t^2 / 1+t^2) \), \( y = 2t / 1+t^2 \)
(S) The eccentricity of the conic lies in the interval \( 1 < x < \infty \)
(T) Points z in the complex plane satisfying \( \operatorname{Re}(z+1)^2 =|z|^2 + 1 \)
Answer:
Solution:
(P) \( \frac{|0 + 0 - 1|}{\sqrt{h^2 + k^2}} = 2 \)
\( h^2 + k^2 = \left(\frac{1}{2}\right)^2 \) which is a circle
(Q) If \( |z - z_1| - |z - z_2| = k \) where \( k < |z_1 - z_2| \) the locus is a hyperbola.
(R) Let t = tan \alpha
\( x = \sqrt{3} \cos 2\alpha \), \( y = \sin 2\alpha \)
\( \frac{x^2}{3} + y^2 = 1 \) Ellipse
(S) e>1 for a hyperbola If e = 1 then parabola
(T) Let z = x + iy
\( (x + 1)^2 - y^2 = x^2 + y^2 + 1 \)
\( y^2 = x \), parabola

Question. An ellipse intersects the hyperbola \( 2x^2 - 2y^2 = 1 \) orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then 
(a) equation of ellipse is \( x^2 + 2y^2 = 2 \)
(b) the foci of ellipse are (±1, 0)
(c) equation of ellipse is \( x^2 + 2y^2 = 4 \)
(d) the foci of ellipse are \( (\pm \sqrt{2} , 0) \)
Answer: (a) equation of ellipse is \( x^2 + 2y^2 = 2 \) and (b) the foci of ellipse are (±1, 0)
Solution:
Ellipse and hyperbola will be confocal
\( \implies \) (± ae, 0) = (± 1, 0)
\( \implies \) \( \left(\pm a \times \frac{1}{\sqrt{2}}, 0\right) \) = (± 1, 0)
\( a = \sqrt{2} \) and \( e = \frac{1}{\sqrt{2}} \)
\( b^2 = a^2 (1 - e^2) \)
\( \implies \) \( b^2 = 1 \)
Ellipse \( \frac{x^2}{2} + \frac{y^2}{1} = 1 \)

Paragraph:
The circle \( x^2 + y^2 - 8x = 0 \) and hyperbola \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \) intersect at the points A and B.

Question. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
(a) \( 2x - \sqrt{5} y - 20 = 0 \)
(b) \( 2x - \sqrt{5} y + 4 = 0 \)
(c) \( 3x - 4y + 8 = 0 \)
(d) \( 4x - 3y + 4 = 0 \)
Answer: (b) \( 2x - \sqrt{5} y + 4 = 0 \)
Solution:
Tangent to ellipse \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \) is
\( y = mx + \sqrt{9m^2 - 4} \), m > 0
It is tangent to \( x^2 + y^2 - 8x = 0 \)
\( \frac{4m + \sqrt{9m^2 - 4}}{\sqrt{1+m^2}} = 4 \)
\( 495 m^4 + 104m^2 - 400 = 0 \)
\( m^2 = \frac{4}{5} \)
\( \implies \) \( m = \frac{2}{\sqrt{5}} \)
Tangent is \( y = \frac{2}{\sqrt{5}} x + \frac{4}{\sqrt{5}} \)
\( \implies \) \( 2x - \sqrt{5} y + 4 = 0 \)

Question. Equation of the circle with AB as its diameter is
(a) \( x^2 + y^2 - 12x + 24 = 0 \)
(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: (a) \( x^2 + y^2 - 12x + 24 = 0 \)
Solution:
A point on hyperbola is \( (3 \sec \theta, 2 \tan \theta) \)
It is lie on circle so
\( 9 \sec^2 \theta + 4 \tan^2 \theta - 24 \sec \theta = 0 \)
\( \implies \) \( 13 \sec^2 \theta - 24 \sec \theta - 4 = 0 \)
\( \implies \) \( \sec \theta = 2, -\frac{2}{13} \)
\( \therefore \sec \theta = 2 \)
\( \implies \) \( \tan \theta = \sqrt{3} \)
POI are \( A(6, 2 \sqrt{3}) \) and \( B(6, -2 \sqrt{3}) \)
equation of circle
\( (x - 6)^2 + y^2 = (2 \sqrt{3})^2 \)
\( \implies \) \( x^2 + y^2 - 12x + 24 = 0 \)

Question. Let the eccentricity of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) be reciprocal to that of the ellipse \( x^2 + 4y^2 = 4 \). If the hyperbola passes through a focus of the ellipse, then
(a) the equation of the hyperbola is \( \frac{x^2}{3} - \frac{y^2}{2} =1 \)
(b) a focus of the hyperbola is (2, 0)
(c) the eccentricity of the hyperbola is \( \sqrt{\frac{5}{3}} \)
(d) the equation of the hyperbola is \( x^2 - 3y^2 = 3 \)
Answer: (b) a focus of the hyperbola is (2, 0) and (d) the equation of the hyperbola is \( x^2 - 3y^2 = 3 \)
Solution:
\( \frac{x^2}{4} + y^2 = 1 \)
\( e = \sqrt{3} / 2 \)
so eccenticity of hyperbola = \( 2 / \sqrt{3} \)
\( \implies \) \( 1 + \frac{b^2}{a^2} = \frac{4}{3} \)
\( \implies \) \( 3b^2 = a^2 \) .....(1)
the equation of H.B. \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \),
passes through \( (\pm \sqrt{3} , 0) \)
\( \implies \) \( b^2 = 1 \)
so equation of hyperbola \( \frac{x^2}{3} - y^2 = 1 \)

Question. Let P(6, 3) be a point on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). If the normal at the point P intersects the x-axis at (9, 0), then the eccentricity of the hyperbola is
(a) \( \sqrt{\frac{5}{2}} \)
(b) \( \sqrt{\frac{3}{2}} \)
(c) \( \sqrt{2} \)
(d) \( \sqrt{3} \)
Answer: (b) \( \sqrt{\frac{3}{2}} \)
Solution:
Equation of normal at P(6, 3)
\( y - 3 = -\frac{3}{6} \cdot \frac{a^2}{b^2} (x - 6) \)
put y = 0,
\( \implies \) \( \frac{6b^2}{a^2} + 6 = 9 \)
\( \implies \) \( 2b^2 = a^2 \)
\( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{3}{2}} \)