JEE Mathematics Hyperbola MCQs Set 08

Answer: (d) lies outside the hyperbola

Explanation:
1. The equation is \( 9(x - y)^2 - 4(x + y)^2 = 36 \), i.e. \( \frac{(x-y)^2}{4} - \frac{(x+y)^2}{9} = 1 \). The axes are \( x - y = 0 \) (conjugate) and \( x + y = 0 \) (transverse); \( (3, 2) \) lies on neither.
2. \( S_1 = (9 - 6)^2 - (6 + 4)^2 - 36 = 9 - 100 - 36 < 0 \).
3. A negative value of the expression in this form means the point lies outside the hyperbola.

Teacher's Note:
1. First check the axis equations, then use the sign of \( S_1 \).
2. Points between the branches give a negative value here.

Answer: (b) \( \sqrt{2} \)

Explanation:
1. Complete the squares: \( (x - 2)^2 - (y - 2)^2 = -16 \).
2. This is a hyperbola with equal semi-axes (a rectangular hyperbola).
3. For a rectangular hyperbola \( e = \sqrt{2} \).

Teacher's Note:
1. Equal coefficients with opposite signs give a rectangular hyperbola.
2. Here \( a = b \), so \( e = \sqrt{1 + 1} \).

Answer: (c) \( \sec\alpha,\ \text{cosec}\,\alpha \)

Explanation:
1. For conjugate hyperbolas \( \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \).
2. Eccentricities of hyperbolas exceed 1, ruling out (a); in (b) \( \tan\alpha \) and \( \cot\alpha \) cannot both exceed 1; in (d) the identity fails.
3. For \( \sec\alpha, \text{cosec}\,\alpha \): \( \cos^2\alpha + \sin^2\alpha = 1 \), so (c) works.

Teacher's Note:
1. Recall \( \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \).
2. Both eccentricities must be greater than 1.

Answer: (a) The length of its transverse axis is \( 4\sqrt{3} \).

Explanation:
1. \( 16(x-1)^2 - 3(y-2)^2 = 44 + 16 - 12 = 48 \), so \( \frac{(x-1)^2}{3} - \frac{(y-2)^2}{16} = 1 \).
2. \( a^2 = 3,\ b^2 = 16 \): transverse axis \( 2a = 2\sqrt{3} \) (not \( 4\sqrt{3} \)), conjugate axis \( 2b = 8 \), centre \( (1, 2) \), \( e^2 = 1 + \frac{16}{3} = \frac{19}{3} \).
3. So only statement (a) is incorrect.

Teacher's Note:
1. Reduce to standard form by completing squares.
2. Check each statement against \( a^2 = 3,\ b^2 = 16 \).

Answer: (c) \( \frac{2}{\sqrt{3}} \)

Explanation:
1. Conjugate axis \( 2b = \frac{1}{2}(2ae) = ae \), so \( b^2 = \frac{a^2e^2}{4} \).
2. Also \( b^2 = a^2(e^2 - 1) \), so \( \frac{e^2}{4} = e^2 - 1 \).
3. \( e^2 = \frac{4}{3} \), so \( e = \frac{2}{\sqrt{3}} \).

Teacher's Note:
1. Distance between the foci is \( 2ae \).
2. Use \( b^2 = a^2(e^2 - 1) \).

Answer: (a) \( \frac{\pi}{4} \)

Explanation:
1. \( OP^2 = 2\sec^2\theta + 2\tan^2\theta = 6 \).
2. \( \sec^2\theta + \tan^2\theta = 3 \), so \( 1 + 2\tan^2\theta = 3 \) and \( \tan\theta = 1 \).
3. In the first quadrant \( \theta = \frac{\pi}{4} \).

Teacher's Note:
1. Use \( \sec^2\theta = 1 + \tan^2\theta \).
2. First quadrant means \( \tan\theta > 0 \).

Answer: (c) \( (4, 1) \)

Explanation:
1. \( 9(x+1)^2 - 16(y-1)^2 = 151 + 9 - 16 = 144 \), so \( \frac{(x+1)^2}{16} - \frac{(y-1)^2}{9} = 1 \).
2. Centre \( (-1, 1) \); \( c^2 = 16 + 9 = 25 \), so \( ae = 5 \).
3. Foci \( (-1 \pm 5, 1) = (4, 1) \) and \( (-6, 1) \).

Teacher's Note:
1. Shift the origin to the centre.
2. The foci are at distance \( ae = \sqrt{a^2 + b^2} \) from the centre along the transverse axis.

Answer: (a) \( (3x - 2)(3y - 2) = 16 \)

Explanation:
1. Let \( B = \left(2t, \frac{2}{t}\right) \). P divides AB with \( AP : PB = 2 : 1 \), so \( P = \frac{A + 2B}{3} \).
2. \( h = \frac{2 + 4t}{3},\ k = \frac{2 + \frac{4}{t}}{3} \), so \( 3h - 2 = 4t \) and \( 3k - 2 = \frac{4}{t} \).
3. Multiplying: \( (3h - 2)(3k - 2) = 16 \).

Teacher's Note:
1. Parametrise the second point of the hyperbola as \( (2t, 2/t) \).
2. Eliminate \( t \) by multiplying the two expressions.

Answer: (a) zero

Explanation:
1. \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \) and the given line has slope \( -\frac{5}{2} \), so the tangent slope is \( m = \frac{2}{5} \).
2. A tangent \( y = mx + c \) needs \( c^2 = a^2m^2 - b^2 = 9\cdot\frac{4}{25} - 4 = \frac{36}{25} - 4 < 0 \).
3. So no real tangent exists with this slope: the number is zero.

Teacher's Note:
1. The slope must exceed \( \frac{b}{a} \) in magnitude for a tangent to exist.
2. Here \( \frac{2}{5} < \frac{2}{3} \).

Answer: (b) \( y = -x \pm 3 \)

Explanation:
1. \( \frac{x^2}{18} - \frac{y^2}{9} = 1 \), and the required slope is \( m = -1 \).
2. \( c^2 = a^2m^2 - b^2 = 18 - 9 = 9 \), so \( c = \pm 3 \).
3. The tangents are \( y = -x \pm 3 \).

Teacher's Note:
1. Perpendicular to \( y = x \) means slope \( -1 \).
2. Use \( c^2 = a^2m^2 - b^2 \).

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

Explanation:
1. For midpoint \( (h, k) \) use \( T = S_1 \): \( 3xh - 2yk + 2(x + h) - 3(y + k) = 3h^2 - 2k^2 + 4h - 6k \).
2. Collect: \( x(3h + 2) + y(-2k - 3) = \ldots \); its slope is \( \frac{3h + 2}{2k + 3} \).
3. Parallel to \( y = 2x \): \( 3h + 2 = 4k + 6 \), so \( 3x - 4y = 4 \).

Teacher's Note:
1. Use the chord with a given midpoint, \( T = S_1 \).
2. Set its slope equal to 2.

Answer: (b) \( 9x - 8y = 72 \)

Explanation:
1. PQ is the chord of contact of \( R(h, k) \): \( \frac{hx}{16} - \frac{ky}{9} = 1 \).
2. It passes through \( (2, 1) \): \( \frac{2h}{16} - \frac{k}{9} = 1 \), i.e. \( \frac{h}{8} - \frac{k}{9} = 1 \).
3. So \( 9h - 8k = 72 \); the locus is \( 9x - 8y = 72 \) (the polar of L).

Teacher's Note:
1. The chord joining the contact points is the polar of the external point.
2. Substitute L into the polar equation.

Answer: (d) infinite

Explanation:
1. The hyperbola is \( \frac{x^2}{\cos^2\alpha} - \frac{y^2}{\sin^2\alpha} = 1 \), with director circle \( x^2 + y^2 = a^2 - b^2 = \cos^2\alpha - \sin^2\alpha = \cos 2\alpha \).
2. For \( 0 < \alpha < \frac{\pi}{4} \): \( 0 < \cos 2\alpha < 1 \), so the director circle is a real circle.
3. Every point of the circle works, so there are infinitely many points.

Teacher's Note:
1. The director circle of a hyperbola is \( x^2 + y^2 = a^2 - b^2 \).
2. It exists only when \( a > b \).

Answer: (a) \( y + mx = 0 \)

Explanation:
1. For midpoint \( (h, k) \), \( T = S_1 \): \( \frac{xk + yh}{2} = hk \), i.e. \( \frac{x}{h} + \frac{y}{k} = 2 \).
2. Its slope is \( m = -\frac{k}{h} \).
3. So \( k = -mh \): the locus is \( y = -mx \), i.e. \( y + mx = 0 \).

Teacher's Note:
1. Find the slope of the chord with midpoint \( (h, k) \).
2. The midpoints of parallel chords lie on a diameter through the centre.

Answer: (d) \( \frac{\pi}{2} \)

Explanation:
1. The hyperbola is \( \frac{x^2}{25} - \frac{y^2}{16} = 1 \) with director circle \( x^2 + y^2 = 25 - 16 = 9 \).
2. \( 1 + 8 = 9 \), so the point lies on the director circle.
3. Tangents from it are perpendicular: the angle is \( \frac{\pi}{2} \).

Teacher's Note:
1. Test whether the point lies on the director circle.
2. Perpendicular tangents come from the director circle.

Answer: (c) a hyperbola with centre \( \left(\frac{\alpha}{2}, \frac{\beta}{2}\right) \)

Explanation:
1. For midpoint \( (h, k) \): \( \frac{xh}{a^2} - \frac{yk}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2} \), passing through \( (\alpha, \beta) \).
2. Rearranging: \( \left(\frac{h^2}{a^2} - \frac{\alpha h}{a^2}\right) - \left(\frac{k^2}{b^2} - \frac{\beta k}{b^2}\right) = 0 \).
3. Completing squares gives \( \frac{(h - \alpha/2)^2}{a^2} - \frac{(k - \beta/2)^2}{b^2} = \frac{\alpha^2}{4a^2} - \frac{\beta^2}{4b^2} \): a hyperbola with centre \( \left(\frac{\alpha}{2}, \frac{\beta}{2}\right) \).

Teacher's Note:
1. Use \( T = S_1 \) and complete the squares.
2. Opposite signs of the squared terms give a hyperbola.