JEE Mathematics Hyperbola MCQs Set 09

Answer: (c) \( \frac{13}{12} \)

Explanation:
1. \( 2b=5 \) gives \( b=\frac52 \); \( 2c=13 \) gives \( c=\frac{13}2 \).
2. \( a^2=c^2-b^2=\frac{169-25}{4}=36 \), so \( a=6 \) and \( e=\frac ca=\frac{13}{12} \).

Teacher's Note:
1. For a hyperbola c² = a² + b², so a² = c² − b².

Answer: (a) a hyperbola

Explanation:
1. Tangency needs \( \beta^2=a^2\alpha^2-b^2 \).
2. Replacing \( (\alpha,\beta) \) by \( (x,y) \): \( a^2x^2-y^2=b^2 \), i.e. \( \frac{x^2}{b^2/a^2}-\frac{y^2}{b^2}=1 \), a hyperbola.

Teacher's Note:
1. Write the tangency condition c² = a²m² − b² with m = α and c = β.

Answer: (c) a hyperbola if \( 8 \lt \lambda \lt 12 \)

Explanation:
1. If \( \lambda \lt 8 \) both denominators are positive and unequal, so it is an ellipse. This rules out (a), and (b) is false because \( \lambda=9 \) gives a hyperbola.
2. If \( 8 \lt \lambda \lt 12 \), \( 12-\lambda \gt 0 \) and \( 8-\lambda \lt 0 \): the signs differ, so it is a hyperbola.

Teacher's Note:
1. The signs of the two denominators decide between an ellipse and a hyperbola.

Answer: (c) \( 2xy-4x+4y+1=0 \)

Explanation:
1. \( SP^2=e^2\cdot(\text{distance to directrix})^2 \): \( (x-1)^2+(y+1)^2=2\cdot\frac{(x-y+1)^2}{2} \).
2. Expanding: \( x^2+y^2-2x+2y+2=x^2+y^2+1-2xy+2x-2y \), which gives \( 2xy-4x+4y+1=0 \).

Teacher's Note:
1. Equate the squared focal distance with e² times the squared distance from the directrix.

Answer: (b) \( b^2 \)

Explanation:
1. For the tangent \( y=mx+c \), \( c^2=a^2m^2-b^2 \).
2. The product of the distances from \( (\pm ae,0) \) is \( \left|\frac{c^2-a^2e^2m^2}{1+m^2}\right|=\frac{b^2(1+m^2)}{1+m^2}=b^2 \).

Teacher's Note:
1. The same result holds for the ellipse and the hyperbola: the product is b².

Answer: (a) \( 3x-4y=4 \)

Explanation:
1. For the mid-point \( (h,k) \), \( T=S_1 \) gives the chord \( 3hx-2ky+2(x+h)-3(y+k)=3h^2-2k^2+4h-6k \) with slope \( \frac{3h+2}{2k+3} \).
2. This slope is 2: \( 3h+2=4k+6 \), i.e. \( 3h-4k=4 \).
3. The locus is \( 3x-4y=4 \).

Teacher's Note:
1. The chord with mid-point (h, k) is T = S₁; equate its slope to 2.

Answer: (d) \( -\left(\frac{a^2+b^2}{b}\right) \)

Explanation:
1. The normal at \( \theta \) is \( ax\cos\theta+\frac{by\cos\theta}{\sin\theta}=a^2+b^2 \). For \( \phi=\frac\pi2-\theta \) the normal is \( ax\sin\theta+\frac{by\sin\theta}{\cos\theta}=a^2+b^2 \).
2. Solving the two equations simultaneously gives \( k=-\frac{a^2+b^2}{b} \) (and \( h \) depends on \( \theta \)).

Teacher's Note:
1. Solve the two normals together; the y-coordinate comes out independent of θ.

Answer: (c) \( a^2y(x-h)+b^2x(y-k)=0 \)

Explanation:
1. The normal at the foot \( (x,y) \) is \( \frac{a^2X}{x}+\frac{b^2Y}{y}=a^2+b^2 \). It passes through \( (h,k) \): \( \frac{a^2h}{x}+\frac{b^2k}{y}=a^2+b^2 \).
2. Multiplying by \( xy \): \( (a^2+b^2)xy-a^2hy-b^2kx=0 \), which is \( a^2y(x-h)+b^2x(y-k)=0 \).

Teacher's Note:
1. Write that the normal at the foot passes through (h, k) and clear the fractions.

Answer: (d) \( \frac{\sqrt{13}+1}{2\sqrt3} \)

Explanation:
1. The ends of the latus rectum are \( \left(ae,\pm\frac{b^2}{a}\right) \); the centre is \( O \). Equal sides: \( a^2e^2+\frac{b^4}{a^2}=\frac{4b^4}{a^2} \), so \( a^4e^2=3b^4 \).
2. With \( b^2=a^2(e^2-1) \): \( e^2=3(e^2-1)^2 \), i.e. \( 3e^4-7e^2+3=0 \) and \( e^2=\frac{7+\sqrt{13}}{6} \) (\( e^2 \gt 1 \)).
3. \( \left(\frac{\sqrt{13}+1}{2\sqrt3}\right)^2=\frac{14+2\sqrt{13}}{12}=\frac{7+\sqrt{13}}{6} \), so \( e=\frac{\sqrt{13}+1}{2\sqrt3} \).

Teacher's Note:
1. Equate the distance from the centre to an end of the latus rectum with the length of the latus rectum.

Answer: (b) \( 1 \)

Explanation:
1. \( e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2} \) and \( e'^2=1+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2} \).
2. \( \frac1{e^2}+\frac1{e'^2}=\frac{a^2+b^2}{a^2+b^2}=1 \).

Teacher's Note:
1. The reciprocals of the squared eccentricities of conjugate hyperbolas always add up to 1.

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

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{16}=1 \): \( a=3 \), \( c=5 \), \( e=\frac53 \).
2. The directrices are \( x=\pm\frac ae=\pm\frac95 \); \( x=-\frac95 \) matches \( 5x+9=0 \).
3. The corresponding focus is \( (-ae,0)=(-5,0) \).

Teacher's Note:
1. The directrix x = −a/e belongs to the focus (−ae, 0).

Answer: (a) \( 5:4 \)

Explanation:
1. A tangent \( y=mx+\frac3m \) to the parabola touches \( x^2-\frac{y^2}{8}=1 \) if \( \frac9{m^2}=m^2-8 \), so \( m^2=9 \).
2. The common tangents are \( y=3x+1 \) and \( y=-3x-1 \), meeting at \( P\left(-\frac13,0\right) \).
3. The foci are \( S(3,0) \) and \( S'(-3,0) \): \( SP=\frac{10}{3} \), \( PS'=\frac83 \), ratio \( 5:4 \).

Teacher's Note:
1. Find the common tangents using the parabola's slope form and the hyperbola's tangency condition.

Answer: (d) \( x-y+4=0 \)

Explanation:
1. A tangent to the parabola is \( y=mx+\frac4m \). Putting it in \( xy=-4 \): \( mx^2+\frac4mx+4=0 \).
2. Tangency needs \( \frac{16}{m^2}-16m=0 \), so \( m=1 \) and the tangent is \( y=x+4 \), i.e. \( x-y+4=0 \).

Teacher's Note:
1. Substitute the parabola's tangent into the second curve and set the discriminant to zero.

Answer: (c) \( 17 \)

Explanation:
1. The tangent \( \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1 \) cuts the axes at \( x=a\cos\theta \) and \( y=-b\cot\theta \).
2. Area \( =\frac12ab\cos\theta\cot\theta=\frac12ab\cdot\frac34\cdot2=\frac{3ab}{4} \) at \( \theta=\frac\pi6 \). Setting it equal to \( 3a^2 \) gives \( b=4a \).
3. \( e^2=1+\frac{b^2}{a^2}=17 \).

Teacher's Note:
1. Find the intercepts of the tangent at the parameter value and use the triangle area formula.

Answer: (d) \( \frac1{a^2}-\frac1{b^2} \)

Explanation:
1. Let \( CP=r \) at angle \( \theta \): \( \frac{r^2\cos^2\theta}{a^2}-\frac{r^2\sin^2\theta}{b^2}=1 \), so \( \frac1{r^2}=\frac{\cos^2\theta}{a^2}-\frac{\sin^2\theta}{b^2} \).
2. For \( CQ \) at \( \theta+90^\circ \): \( \frac1{CQ^2}=\frac{\sin^2\theta}{a^2}-\frac{\cos^2\theta}{b^2} \).
3. Adding: \( \frac1{CP^2}+\frac1{CQ^2}=\frac1{a^2}-\frac1{b^2} \).

Teacher's Note:
1. Use polar form from the centre for P and for the perpendicular direction, then add.

Answer: (d) \( (4,-\sqrt6) \)

Explanation:
1. The tangent at \( (x_1,y_1) \) is \( \frac{xx_1}{4}-\frac{yy_1}{2}=1 \). Compare with \( x+\frac{\sqrt6}{2}y=1 \).
2. \( \frac{x_1}{4}=1 \) gives \( x_1=4 \); \( -\frac{y_1}{2}=\frac{\sqrt6}{2} \) gives \( y_1=-\sqrt6 \). Check: \( \frac{16}{4}-\frac62=1 \).

Teacher's Note:
1. Compare the line with the tangent at (x₁, y₁) and verify the point lies on the curve.

Answer: (b) \( \sqrt{\frac32}-1 \)

Explanation:
1. Completing squares: \( (x-\sqrt2)^2-2(y+\sqrt2)^2=4 \), i.e. \( \frac{X^2}{4}-\frac{Y^2}{2}=1 \) with \( a=2 \), \( b=\sqrt2 \), \( c=\sqrt6 \).
2. Taking the branch on the right: \( A=(2,0) \), nearest focus \( C=(\sqrt6,0) \), \( B=\left(\sqrt6,\frac{b^2}{a}\right)=(\sqrt6,1) \) (relative to the centre).
3. Area \( =\frac12(\sqrt6-2)(1)=\sqrt{\frac32}-1 \).

Teacher's Note:
1. Shift to the centre, read off a, b, c, then use base AC and height equal to the semi-latus rectum.

Answer: (c) \( \frac{\sqrt{13}}{3} \)

Explanation:
1. The lines \( 2y-x-3=0 \) and \( 2x+y-1=0 \) are perpendicular, each with \( \sqrt{1+4}=\sqrt5 \). Put \( X=\frac{2y-x-3}{\sqrt5} \) and \( Y=\frac{2x+y-1}{\sqrt5} \).
2. Then \( 20X^2-45Y^2=80 \), i.e. \( \frac{X^2}{4}-\frac{Y^2}{16/9}=1 \).
3. \( e^2=1+\frac{16/9}{4}=\frac{13}{9} \), so \( e=\frac{\sqrt{13}}{3} \).

Teacher's Note:
1. Rotate to perpendicular axes by normalising the two lines.

Answer: (d) \( e \gt \frac{2}{\sqrt3} \)

Explanation:
1. Take \( A(x_1,y_1) \), \( B(x_1,-y_1) \) with \( y_1=\frac{x_1}{\sqrt3} \) (angle \( 30^\circ \) at \( O \)). On the curve: \( x_1^2\left(\frac1{a^2}-\frac1{3b^2}\right)=1 \).
2. A real solution needs \( 3b^2 \gt a^2 \), i.e. \( 3(e^2-1) \gt 1 \) and \( e \gt \frac2{\sqrt3} \).

Teacher's Note:
1. A real point exists only if the bracket is positive, which restricts e.

Answer: (c) \( a^2 \)

Explanation:
1. The lines \( x\pm y=0 \) are the asymptotes. The tangent at \( (a\sec\theta,a\tan\theta) \) is \( x\sec\theta-y\tan\theta=a \).
2. It meets the asymptotes at \( \left(\frac{a}{\sec\theta-\tan\theta},\frac{a}{\sec\theta-\tan\theta}\right) \) and \( \left(\frac{a}{\sec\theta+\tan\theta},-\frac{a}{\sec\theta+\tan\theta}\right) \), so the area is \( \frac12\cdot\sqrt2\cdot\frac{a}{\sec\theta-\tan\theta}\cdot\sqrt2\cdot\frac{a}{\sec\theta+\tan\theta}=a^2 \).

Teacher's Note:
1. For any hyperbola, the triangle cut off from the asymptotes by a tangent has constant area ab.