JEE Mathematics Hyperbola MCQs Set 10

Answer: (a) \( (6,5\sqrt2) \)

Explanation:
1. \( a=2 \), \( c=3 \), so \( b^2=5 \) and the hyperbola is \( \frac{x^2}{4}-\frac{y^2}{5}=1 \).
2. \( (6,5\sqrt2) \): \( 9-10=-1\neq1 \). The other three points give \( 6-5=1 \), \( 9-8=1 \) and \( 4-3=1 \).
3. So \( (6,5\sqrt2) \) does not lie on it.

Teacher's Note:
1. Substitute each point into the equation obtained from a and c.

Answer: (d) \( x-y+1=0 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{5}-\frac{y^2}{4}=1 \). For slope \( m=1 \): \( c^2=a^2m^2-b^2=5-4=1 \).
2. The tangents are \( y=x\pm1 \), i.e. \( x-y+1=0 \) or \( x-y-1=0 \). Only \( x-y+1=0 \) is listed.

Teacher's Note:
1. Use the tangency condition c² = a²m² − b² for a hyperbola.

Answer: (a) \( (2\sqrt2,3\sqrt3) \)

Explanation:
1. With \( c=2 \): \( a^2+b^2=4 \). Through \( P \): \( \frac{2}{a^2}-\frac{3}{4-a^2}=1 \), giving \( a^4-9a^2+8=0 \), so \( a^2=1 \) (as \( a^2 \lt 4 \)) and \( b^2=3 \).
2. The hyperbola is \( x^2-\frac{y^2}{3}=1 \) and the tangent at \( P \) is \( \sqrt2x-\frac{\sqrt3y}{3}=1 \).
3. \( (2\sqrt2,3\sqrt3) \): \( 4-3=1 \). The other points fail.

Teacher's Note:
1. Find a² and b² from the focus and the point, then write the tangent at the point.

Answer: (d) \( 45\sqrt5 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{36}=1 \). The chord of contact from \( (0,3) \) is \( -\frac{3y}{36}=1 \), i.e. \( y=-12 \).
2. Putting \( y=-12 \): \( 4x^2=36+144 \), \( x=\pm3\sqrt5 \), so \( PQ=6\sqrt5 \).
3. The distance of \( T \) from the chord is \( 15 \), so the area is \( \frac12\cdot6\sqrt5\cdot15=45\sqrt5 \).

Teacher's Note:
1. Use the chord of contact from the external point, then base times height.

Answer: (c) \( y=\frac6x \)

Explanation:
1. The tangent at \( (x,y) \) meets the axes at \( (2x,0) \) and \( (0,2y) \), so its slope is \( -\frac{y}{x} \), i.e. \( \frac{dy}{dx}=-\frac yx \).
2. Integrating gives \( xy=k \); through \( (2,3) \), \( k=6 \), so \( y=\frac6x \).

Teacher's Note:
1. Midpoint condition on the intercepts gives the differential equation dy/dx = −y/x.

Answer: (c) \( 3x^2-y^2=3 \)

Explanation:
1. \( c=2 \) and \( e=2 \) give \( a=1 \), \( b^2=c^2-a^2=3 \).
2. The hyperbola is \( x^2-\frac{y^2}{3}=1 \), i.e. \( 3x^2-y^2=3 \).

Teacher's Note:
1. a = c/e and b² = c² − a².

Answer: (b) \( \frac{(x-5)^2}{25}-\frac{y^2}{144}=1 \)

Explanation:
1. The centre is the mid-point of the vertices, \( (5,0) \), and \( a=5 \).
2. \( c=18-5=13 \), so \( b^2=169-25=144 \).
3. The equation is \( \frac{(x-5)^2}{25}-\frac{y^2}{144}=1 \).

Teacher's Note:
1. Find the centre first, then a and c as distances from it.

Answer: (b) \( a^2\cos^2\alpha-b^2\sin^2\alpha=p^2 \)

Explanation:
1. Write the line as \( y=-\cot\alpha\,x+\frac{p}{\sin\alpha} \). Then \( c^2=a^2m^2-b^2 \) gives \( \frac{p^2}{\sin^2\alpha}=a^2\cot^2\alpha-b^2 \).
2. Multiplying by \( \sin^2\alpha \): \( p^2=a^2\cos^2\alpha-b^2\sin^2\alpha \).

Teacher's Note:
1. For the ellipse the sign is plus, for the hyperbola it is minus.

Answer: (c) \( a^2-b^2=2c^2 \)

Explanation:
1. Curves cut at right angles when they are confocal. The hyperbola \( \frac{x^2}{c^2}-\frac{y^2}{c^2}=1 \) has foci at \( \pm c\sqrt2 \).
2. The ellipse has foci at \( \pm\sqrt{a^2-b^2} \), so \( a^2-b^2=2c^2 \).

Teacher's Note:
1. Orthogonal intersection of an ellipse and a hyperbola happens when they have the same foci.

Answer: (a) \( 2\tan^{-1}\left(\frac ba\right) \)

Explanation:
1. The asymptotes are \( y=\pm\frac bax \), each inclined at \( \tan^{-1}\frac ba \) to the x-axis.
2. The angle between them (containing the hyperbola) is \( 2\tan^{-1}\frac ba \).

Teacher's Note:
1. Each asymptote makes angle tan⁻¹(b/a) with the transverse axis.

Answer: (a) \( (3,\infty) \)

Explanation:
1. \( e^2=1+\tan^2\theta=\sec^2\theta \gt 4 \), so \( \cos\theta \lt \frac12 \), i.e. \( \theta\in\left(\frac\pi3,\frac\pi2\right) \).
2. Latus rectum \( =\frac{2\sin^2\theta}{\cos\theta} \), which increases with \( \theta \): at \( \frac\pi3 \) it is \( \frac{2\cdot3/4}{1/2}=3 \) and it tends to infinity.
3. So it lies in \( (3,\infty) \).

Teacher's Note:
1. Convert e > 2 into a range of θ, then see how the latus rectum behaves on it.

Answer: (b) \( 2 \)

Explanation:
1. From the first line \( t=\frac{\sqrt3x-y}{4\sqrt3} \); the second gives \( t(\sqrt3x+y)=4\sqrt3 \).
2. Eliminating \( t \): \( (\sqrt3x-y)(\sqrt3x+y)=48 \), i.e. \( 3x^2-y^2=48 \) or \( \frac{x^2}{16}-\frac{y^2}{48}=1 \).
3. \( e^2=1+\frac{48}{16}=4 \), so \( e=2 \).

Teacher's Note:
1. Eliminate the parameter by multiplying the two expressions for t.

Answer: (b) \( \frac43 \)

Explanation:
1. \( a=2 \). Through \( (4,2) \): \( 4-\frac{4}{b^2}=1 \), so \( b^2=\frac43 \).
2. Latus rectum \( =\frac{2b^2}{a}=\frac{2\cdot4/3}{2}=\frac43 \).

Teacher's Note:
1. Latus rectum of a hyperbola is 2b²/a.

Answer: (c) \( \frac2{\sqrt5} \)

Explanation:
1. The line \( y=mx+c \) is a normal if \( c^2=\frac{m^2(a^2+b^2)^2}{a^2-b^2m^2} \).
2. \( 147(24-18m^2)=42^2m^2 \) gives \( 3528=4410m^2 \), so \( m^2=\frac45 \) and \( m=\pm\frac2{\sqrt5} \).

Teacher's Note:
1. Use the condition for y = mx + c to be a normal to the hyperbola.

Answer: (a) \( 4e^4-24e^2+35=0 \)

Explanation:
1. Directrix \( \frac ae=\frac{4}{\sqrt5} \) gives \( a^2=\frac{16e^2}{5} \). Also \( b^2=a^2(e^2-1) \).
2. Through \( (4,2\sqrt3) \): \( \frac{16}{a^2}-\frac{12}{a^2(e^2-1)}=1 \), so \( a^2=16-\frac{12}{e^2-1} \).
3. Equating: \( \frac{16e^2}{5}=\frac{16e^2-28}{e^2-1} \), which simplifies to \( 4e^4-24e^2+35=0 \).

Teacher's Note:
1. Express a² through e from the directrix, and again from the point; equate the two.

Answer: (b) \( (2x+3y-8)(x+2y-5)=40 \)

Explanation:
1. Asymptotes through the centre: \( 2x+3y-8=0 \) and \( x+2y-5=0 \). The hyperbola is \( (2x+3y-8)(x+2y-5)=k \).
2. Through \( (2,4) \): \( (4+12-8)(2+8-5)=8\cdot5=40 \).

Teacher's Note:
1. A hyperbola differs from the pair of its asymptotes only by a constant.

Answer: (b) \( 16 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{144/25}-\frac{y^2}{81/25}=1 \), so \( c^2=\frac{144+81}{25}=9 \).
2. For the ellipse \( c^2=25-b^2=9 \), so \( b^2=16 \).

Teacher's Note:
1. Equate c² of the two conics.

Answer: (c) \( \frac1e \)

Explanation:
1. The asymptotes are inclined at \( \tan^{-1}\frac ba \) to the transverse axis, so \( \frac\theta2=\tan^{-1}\frac ba \).
2. \( \cos\frac\theta2=\frac{a}{\sqrt{a^2+b^2}}=\frac1e \).

Teacher's Note:
1. Half the angle between the asymptotes has tangent b/a, so its cosine is a/c = 1/e.

Answer: (b) \( y_1+y_2+y_3+y_4=0 \)

Explanation:
1. Substituting \( y=\frac{c^2}{x} \): \( x^4-a^2x^2+c^4=0 \), so \( \sum x_i=0 \) and \( \prod x_i=c^4 \).
2. By symmetry \( y \) satisfies \( y^4-a^2y^2+c^4=0 \): \( \sum y_i=0 \) and \( \prod y_i=c^4 \). Only (b) is correct.

Teacher's Note:
1. Eliminate one variable to get a quartic and use the relations between roots and coefficients.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. Hyperbola: \( \frac{x^2}{5}-\frac{y^2}{5\cos^2\alpha}=1 \), so \( e_h^2=1+\cos^2\alpha \). Ellipse: \( \frac{x^2}{25\cos^2\alpha}+\frac{y^2}{25}=1 \), so \( e_e^2=1-\cos^2\alpha=\sin^2\alpha \).
2. \( e_h^2=3e_e^2 \): \( 1+\cos^2\alpha=3\sin^2\alpha \), so \( 4\cos^2\alpha=2 \) and \( \alpha=\frac\pi4 \).

Teacher's Note:
1. Write each conic in standard form to read its eccentricity in terms of α.