JEE Mathematics Hyperbola MCQs Set 06

Answer: (c) \( 2x-y-2=0 \)

Explanation:
1. \( e=2 \) gives \( b^2=3a^2 \); passing through \( (4,6) \): \( \dfrac{16}{a^2}-\dfrac{36}{3a^2}=1 \), so \( a^2=4 \), \( b^2=12 \).
2. Tangent: \( \dfrac{4x}4-\dfrac{6y}{12}=1 \), i.e. \( 2x-y-2=0 \). Answer: (c)

Teacher's Note:
1. Use \( e^2=1+b^2/a^2 \).

Answer: (c) \( \dfrac2{\sqrt5} \)

Explanation:
1. A normal of slope \( m \) has \( c^2=\dfrac{m^2(a^2+b^2)^2}{a^2-b^2m^2} \), so \( 147=\dfrac{1764m^2}{24-18m^2} \).
2. This gives \( 3528-2646m^2=1764m^2 \), so \( m^2=\dfrac45 \) and \( m=\pm\dfrac2{\sqrt5} \). Answer: (c)

Teacher's Note:
1. Use the condition for a line to be a normal to a hyperbola.

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

Explanation:
1. The directrix \( x=\dfrac ae \) gives \( a^2=\dfrac{16e^2}5 \), and \( b^2=a^2(e^2-1)=\dfrac{16e^2(e^2-1)}5 \).
2. Passing through the point: \( \dfrac{16}{a^2}-\dfrac{12}{b^2}=1 \), i.e. \( \dfrac5{e^2}-\dfrac{60}{16e^2(e^2-1)}=1 \). Clearing denominators gives \( 16e^4-96e^2+140=0 \), i.e. \( 4e^4-24e^2+35=0 \). Answer: (d)

Teacher's Note:
1. Express \( a \) and \( b \) in terms of \( e \) and substitute the point.

Answer: (b) \( (-5,0) \)

Explanation:
1. The hyperbola is \( \dfrac{x^2}9-\dfrac{y^2}{16}=1 \) with \( e=\dfrac53 \), so the directrices are \( x=\pm\dfrac ae=\pm\dfrac95 \) and the foci are \( (\pm5,0) \).
2. The directrix \( x=-\dfrac95 \) corresponds to the focus \( (-5,0) \). Answer: (b)

Teacher's Note:
1. Directrix \( x=-a/e \) goes with focus \( (-ae,0) \).

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

Explanation:
1. \( e^2=1+\tan^2\theta\gt4 \) gives \( \tan\theta\gt\sqrt3 \), so \( \theta\in\left(\dfrac\pi3,\dfrac\pi2\right) \).
2. Latus rectum \( =\dfrac{2b^2}a=2\sin\theta\tan\theta \), which is increasing on this interval, so it is greater than \( 2\cdot\dfrac{\sqrt3}2\cdot\sqrt3=3 \) and unbounded above. Answer: (b)

Teacher's Note:
1. Both \( \sin\theta \) and \( \tan\theta \) increase on \( (0,\pi/2) \).

Answer: (b) \( \dfrac2{\sqrt3} \)

Explanation:
1. \( 2a=4 \) gives \( a=2 \), and the point \( (4,2) \) gives \( \dfrac{16}4-\dfrac4{b^2}=1 \), so \( b^2=\dfrac43 \).
2. \( e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac13}=\dfrac2{\sqrt3} \). Answer: (b)

Teacher's Note:
1. Use \( e^2=1+b^2/a^2 \).

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

Explanation:
1. The hyperbola is \( \dfrac{x^2}5-\dfrac{y^2}4=1 \), and the tangent has slope \( 1 \): \( y=x\pm\sqrt{a^2m^2-b^2}=x\pm\sqrt{5-4}=x\pm1 \).
2. So the tangents are \( x-y\pm1=0 \), and the option is \( x-y+1=0 \). Answer: (c)

Teacher's Note:
1. Use \( y=mx\pm\sqrt{a^2m^2-b^2} \).

Answer: (a) \( \dfrac{13}{12} \)

Explanation:
1. \( 2b=5 \) and \( 2ae=13 \); with \( b^2=a^2(e^2-1)=a^2e^2-a^2 \): \( \dfrac{25}4=\dfrac{169}4-a^2 \), so \( a^2=36 \), \( a=6 \).
2. \( e=\dfrac{13}{2a}=\dfrac{13}{12} \). Answer: (a)

Teacher's Note:
1. Use \( b^2=a^2e^2-a^2 \).

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

Explanation:
1. \( a=2 \) and \( ae=3 \), so \( b^2=9-4=5 \) and the hyperbola is \( \dfrac{x^2}4-\dfrac{y^2}5=1 \).
2. \( (2\sqrt6,5) \): \( 6-5=1 \); \( (4,\sqrt{15}) \): \( 4-3=1 \); \( (-6,2\sqrt{10}) \): \( 9-8=1 \); but \( (6,5\sqrt2) \): \( 9-10=-1\ne1 \). Answer: (b)

Teacher's Note:
1. Substitute each point in the equation.

Answer: (a) \( 45\sqrt5 \)

Explanation:
1. \( PQ \) is the chord of contact of \( T(0,3) \): \( -\dfrac{3y}{36}=1 \), so \( y=-12 \). Then \( x^2=9+36=45 \), \( x=\pm3\sqrt5 \), giving \( PQ=6\sqrt5 \).
2. The height from \( T \) to \( PQ \) is \( 15 \), so the area is \( \dfrac12\cdot6\sqrt5\cdot15=45\sqrt5 \). Answer: (a)

Teacher's Note:
1. The chord of contact is the polar of \( T \).

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

Explanation:
1. \( ae=2 \) gives \( a^2+b^2=4 \). Through \( P \): \( \dfrac2{a^2}-\dfrac3{4-a^2}=1 \), so \( a^4-9a^2+8=0 \) and \( a^2=1 \) (\( a^2=8 \) is impossible as \( a^2\lt4 \)), \( b^2=3 \).
2. The tangent at \( P \) is \( \sqrt2x-\dfrac{\sqrt3y}3=1 \); the point \( (2\sqrt2,3\sqrt3) \) gives \( 4-3=1 \). Answer: (b)

Teacher's Note:
1. Use \( a^2+b^2=c^2 \) with \( c=2 \).

Answer: (c) \( \dfrac2{\sqrt3} \)

Explanation:
1. \( \dfrac{2b^2}a=8 \) gives \( b^2=4a \), and \( 2b=ae \) gives \( 4b^2=a^2e^2 \).
2. Since \( b^2=a^2(e^2-1) \), \( 4a^2(e^2-1)=a^2e^2 \), so \( 3e^2=4 \) and \( e=\dfrac2{\sqrt3} \). Answer: (c)

Teacher's Note:
1. Use \( b^2=a^2(e^2-1) \).

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

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

Teacher's Note:
1. Foci on the x-axis give a hyperbola opening left and right.

Answer: (d) Abscissa of foci

Explanation:
1. \( a^2=\cos^2\alpha \), \( b^2=\sin^2\alpha \), so \( e^2=1+\tan^2\alpha=\sec^2\alpha \) and \( e=\sec\alpha \).
2. The foci are \( (\pm ae,0)=(\pm1,0) \), so the abscissa of the foci is constant, while \( e=\sec\alpha \) and the vertices \( (\pm\cos\alpha,0) \) vary (and so do the directrices \( x=\pm\cos^2\alpha \)). Answer: (d)

Teacher's Note:
1. \( ae=\cos\alpha\sec\alpha=1 \).

Answer: (a) a hyperbola

Explanation:
1. Tangency of \( y=\alpha x+\beta \) requires \( \beta^2=a^2\alpha^2-b^2 \).
2. So the locus of \( (\alpha,\beta) \) is \( a^2x^2-y^2=b^2 \), a hyperbola. Answer: (a)

Teacher's Note:
1. Apply the tangency condition \( c^2=a^2m^2-b^2 \).

Answer: (a) \( \dfrac x{x_1+x_2}+\dfrac y{y_1+y_2}=1 \)

Explanation:
1. Using \( x_1y_1=x_2y_2=c^2 \), test the line \( \dfrac x{x_1+x_2}+\dfrac y{y_1+y_2}=1 \) at \( (x_1,y_1) \): since \( y_i=c^2/x_i \), \( \dfrac{y_1}{y_1+y_2}=\dfrac{x_2}{x_1+x_2} \), so the left side is \( \dfrac{x_1+x_2}{x_1+x_2}=1 \).
2. By symmetry the line also passes through \( (x_2,y_2) \), so it is the required chord. Answer: (a)

Teacher's Note:
1. Verify that both points satisfy the line.