JEE Mathematics Hyperbola MCQs Set 03

Question. The eccentricity of the hyperbola \( 4x^2 - 9y^2 - 8x = 32 \) is
(a) \( \frac{\sqrt{5}}{3} \)
(b) \( \frac{\sqrt{13}}{3} \)
(c) \( \frac{\sqrt{13}}{9} \)
(d) \( \frac{3}{2} \)
Solution:
\( 4x^2 - 9y^2 - 8x = 32 \)
\( \implies 4(x - 1)^2 - 9y^2 = 36 \)
\( \implies \frac{(x - 1)^2}{9} - \frac{y^2}{4} = 1 \)
\( e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{4}{9} = \frac{13}{9} \)
\( \implies e = \frac{\sqrt{13}}{3} \)
Answer: (b) \( \frac{\sqrt{13}}{3} \)

Question. The locus of the point of intersection of the lines \( \sqrt{3}x - y - 4\sqrt{3}k = 0 \) and \( \sqrt{3}kx + ky - 4\sqrt{3} = 0 \) for different values of k is
(a) ellipse
(b) parabola
(c) circle
(d) hyperbola
Solution:
\( \sqrt{3}x - y - 4\sqrt{3}k = 0 \) ...(1)
\( \sqrt{3}kx + ky - 4\sqrt{3} = 0 \) ...(2)
Solve (1) and (2)
\( x = \frac{2(1+k^2)}{k} \) and \( y = \frac{2\sqrt{3}(1-k^2)}{k} \)
\( \frac{x^2}{4} - \frac{y^2}{12} = 4 \)
\( \implies \frac{x^2}{16} - \frac{y^2}{48} = 1 \) Hyperbola
Answer: (d) hyperbola

Question. If the latus rectum of an hyperbola be 8 and eccentricity be \( \frac{3}{\sqrt{5}} \) then the equation of the hyperbola is
(a) \( 4x^2 - 5y^2 = 100 \)
(b) \( 5x^2 - 4y^2 = 100 \)
(c) \( 4x^2 + 5y^2 = 100 \)
(d) \( 5x^2 + 4y^2 = 100 \)
Solution:
\( \frac{2b^2}{a} = 8 \); \( e = \frac{3}{\sqrt{5}} \)
\( \implies b^2 = 4a \); \( e^2 = \frac{9}{5} \)
\( 1 + \frac{b^2}{a^2} = \frac{9}{5} \)
\( \implies \frac{b^2}{a^2} = \frac{4}{5} \)
\( \implies a = 5 \)
\( \implies b^2 = 20 \)
Hyp. \( \frac{x^2}{25} - \frac{y^2}{20} = 1 \)
\( \implies 4x^2 - 5y^2 = 100 \)
Answer: (a) \( 4x^2 - 5y^2 = 100 \)

Question. If the centre, vertex and focus of a hyperbola be (0, 0), (4, 0) and (6, 0) respectively, then the equation of the hyperbola is
(a) \( 4x^2 - 5y^2 = 8 \)
(b) \( 4x^2 - 5y^2 = 80 \)
(c) \( 5x^2 - 4y^2 = 80 \)
(d) \( 5x^2 - 4y^2 = 8 \)
Solution:
C(0, 0) \( A_1(4, 0) \) \( F_1(6, 0) \)
\( CA_1 = 4 \)
\( CF_1 = 6 \)
\( \implies a = 4 \)
\( ae = 6 \)
\( a^2e^2 = 36 \)
\( \implies a^2\left(1 + \frac{b^2}{a^2}\right) = 36 \)
\( \implies b^2 = 36 - 16 \)
\( \implies b^2 = 20 \)
Hyp. \( \frac{x^2}{16} - \frac{y^2}{20} = 1 \) or \( 5x^2 - 4y^2 = 80 \)
Answer: (c) \( 5x^2 - 4y^2 = 80 \)

Question. The equation of the hyperbola whose foci are (6, 5), (-4, 5) and eccentricity 5/4 is
(a) \( \frac{(x-1)^2}{16} - \frac{(y-5)^2}{9} = 1 \)
(b) \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \)
(c) \( \frac{(x-1)^2}{16} - \frac{(y-5)^2}{9} = -1 \)
(d) None of the options
Solution:
\( F_1(6, 5) \) \( F_2(-4, 5) \) \( e = \frac{5}{4} \)
\( F_1F_2 = 2ae \) Centre of hyp. is the mid point of \( F_1F_2 = (1, 5) \)
\( 2ae = 10 \)
\( \implies ae = 5 \)
\( \implies a^2e^2 = 25 \)
\( \implies a^2\left(\frac{25}{16}\right) = 25 \)
\( \implies a^2 = 16 \)
\( \implies b^2 = 9 \)
Hyp. \( \frac{(x-1)^2}{16} - \frac{(y-5)^2}{9} = 1 \)
Answer: (a) \( \frac{(x-1)^2}{16} - \frac{(y-5)^2}{9} = 1 \)

Question. The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola is
(a) \( \frac{x^2}{25} - \frac{y^2}{144} = 1 \)
(b) \( \frac{(x-5)^2}{25} - \frac{y^2}{144} = 1 \)
(c) \( \frac{x^2}{25} - \frac{(y-5)^2}{144} = 1 \)
(d) \( \frac{(x-5)^2}{25} - \frac{(y-5)^2}{144} = 1 \)
Solution:
Centre of hyp. will be mid point of \( A_1 \) & \( A_2 = \left(\frac{10+0}{2}, 0\right) = (5, 0) \)
Answer: (b) \( \frac{(x-5)^2}{25} - \frac{y^2}{144} = 1 \)

Question. The length of the transverse axis of a hyperbola is 7 and it passes through the point (5, -2). The equation of the hyperbola is
(a) \( \frac{4}{49}x^2 - \frac{196}{51}y^2 = 1 \)
(b) \( \frac{49}{4}x^2 - \frac{51}{196}y^2 = 1 \)
(c) \( \frac{4}{49}x^2 - \frac{51}{196}y^2 = 1 \)
(d) None of the options
Solution:
\( 2a = 7 \)
\( \implies a = \frac{7}{2} \)
Let the Equation of hyp. \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
passes through (5, -2)
\( \frac{25}{a^2} - \frac{4}{b^2} = 1 \)
\( \frac{25}{a^2} - 1 = \frac{4}{b^2} \)
\( b^2 = \frac{4}{\frac{25}{a^2}-1} = \frac{4a^2}{25-a^2} = \frac{4 \times \frac{49}{4}}{25 - \frac{49}{4}} = \frac{196}{51} \)
Equation \( \frac{4x^2}{49} - \frac{51y^2}{196} = 1 \)
Answer: (c) \( \frac{4}{49}x^2 - \frac{51}{196}y^2 = 1 \)

Question. If the eccentricity of the hyperbola \( x^2 - y^2\sec^2\alpha = 5 \) is \( \sqrt{3} \) times the eccentricity of the ellipse \( x^2\sec^2\alpha + y^2 = 25 \), then a value of \( \alpha \) is
(a) \( \pi/6 \)
(b) \( \pi/4 \)
(c) \( \pi/3 \)
(d) \( \pi/2 \)
Solution:
\( x^2 - y^2\sec^2\alpha = 5 \)
\( \frac{x^2}{5} - \frac{y^2}{5\cos^2\alpha} = 1 \rightarrow e_1 \)
\( e_1 = \sqrt{1 + \frac{5\cos^2\alpha}{5}} = \sqrt{1 + \cos^2\alpha} \) ...(1)
\( x^2\sec^2\alpha + y^2 = 25 \)
\( \frac{x^2}{25\cos^2\alpha} + \frac{y^2}{25} = 1 \rightarrow e_2 \)
\( e_2 = \sqrt{1 - \frac{25\cos^2\alpha}{25}} = \sqrt{1 - \cos^2\alpha} \)
\( e_1 = \sqrt{3} e_2 \)
\( e_1^2 = 3e_2^2 \)
\( 1 + \cos^2\alpha = 3 - 3\cos^2\alpha \)
\( 4\cos^2\alpha = 2 \)
\( \cos\alpha = \frac{1}{\sqrt{2}} \)
\( \implies \alpha = \frac{\pi}{4} \)
Answer: (b) \( \pi/4 \)

Question. AB is a double ordinate of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) such that \( \Delta AOB \) (where 'O' is the origin) is an equilateral triangle, then the eccentricity e of the hyperbola satisfies
(a) \( e > \sqrt{3} \)
(b) \( 1 < e < \frac{2}{\sqrt{3}} \)
(c) \( e = \frac{2}{\sqrt{3}} \)
(d) \( e > \frac{2}{\sqrt{3}} \)
Solution:
\( \theta = 30^\circ \)
\( \frac{b\tan\theta}{a\sec\theta} = \tan30^\circ \)
\( \frac{b}{a}\sin\theta = \frac{1}{\sqrt{3}} \)
\( \frac{b}{a} = \frac{1}{\sqrt{3}\sin\theta} \)
\( e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1}{3\sin^2\theta} \)
\( e^2 > 1 + \frac{1}{3} \)
\( e > \frac{2}{\sqrt{3}} \)
Answer: (d) \( e > \frac{2}{\sqrt{3}} \)

Question. The equation of the tangent lines to the hyperbola \( x^2 - 2y^2 = 18 \) which are perpendicular to the line \( y = x \) are
(a) \( y = x \pm 3 \)
(b) \( y = -x \pm 3 \)
(c) \( 2x + 3y + 4 = 0 \)
(d) None of the options
Solution:
\( \frac{x^2}{18} - \frac{y^2}{9} = 1 \)
Slope of tangent would be = -1
\( y = mx \pm \sqrt{a^2m^2 - b^2} \)
\( y = -x \pm \sqrt{18(-1)^2 - 9} \)
\( y = -x \pm 3 \)
Answer: (b) \( y = -x \pm 3 \)

Question. The equation to the common tangents to the two hyperbolas \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) are
(a) \( y = \pm x \pm \sqrt{b^2 - a^2} \)
(b) \( y = \pm x \pm (a^2 - b^2) \)
(c) \( y = \pm x \pm \sqrt{a^2 - b^2} \)
(d) \( y = \pm x \pm \sqrt{a^2 + b^2} \)
Solution:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
Tangent
\( y = mx \pm \sqrt{a^2m^2 - b^2} \) ...(1)
\( \frac{x^2}{(-b^2)} - \frac{y^2}{(-a^2)} = 1 \)
\( y = mx \pm \sqrt{(-b^2)m^2 + a^2} \) ...(2)
(1) and (2) are same
\( \frac{1}{1} = \frac{\sqrt{a^2m^2 - b^2}}{\sqrt{a^2 - b^2m^2}} \)
\( a^2 - b^2m^2 = a^2m^2 - b^2 \)
\( m^2 = 1 \)
\( \implies m = \pm 1 \)
\( y = \pm x \pm \sqrt{a^2 - b^2} \)
Answer: (c) \( y = \pm x \pm \sqrt{a^2 - b^2} \)

Question. Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola \( 16y^2 - 9x^2 = 1 \) is
(a) \( x^2 + y^2 = 9 \)
(b) \( x^2 + y^2 = 1/9 \)
(c) \( x^2 + y^2 = 7/144 \)
(d) \( x^2 + y^2 = 1/16 \)
Solution:
Locus of the feet of the \(\perp^n\) drawn from any focus of the the hyp. upon any tangent is its auxilary circle
Hyp. \( -\frac{x^2}{\left(\frac{1}{9}\right)} + \frac{y^2}{\left(\frac{1}{16}\right)} = 1 \)
Auxiliary circle \( x^2 + y^2 = \frac{1}{16} \)
Answer: (d) \( x^2 + y^2 = 1/16 \)

Question. The ellipse \( 4x^2 + 9y^2 = 36 \) and the hyperbola \( 4x^2 - y^2 = 4 \) have the same foci and they intersect at right angles then the equation of the circle through the points of intersection of two conics is
(a) \( x^2 + y^2 = 5 \)
(b) \( \sqrt{5}(x^2+y^2) - 3x - 4y = 0 \)
(c) \( \sqrt{5}(x^2+y^2) + 3x + 4y = 0 \)
(d) \( x^2 + y^2 = 25 \)
Solution:
If they intersect at right angles then circle will pass through its focus
Circle will be
\( x^2 + y^2 = (OF_1)^2 \)
\( x^2 + y^2 = (\sqrt{5})^2 \)
\( x^2 + y^2 = (\sqrt{5})^2 \); \( F_1(ae, 0) \) \( e = \sqrt{5} \)
\( x^2 + y^2 = 5 \); \( F_1(\sqrt{5}, 0) \)
Answer: (a) \( x^2 + y^2 = 5 \)

Question. The equation of the common tangent to the parabola \( y^2 = 8x \) and the hyperbola \( 3x^2 - y^2 = 3 \) is
(a) \( 2x \pm y + 1 = 0 \)
(b) \( x \pm y + 1 = 0 \)
(c) \( x \pm 2y + 1 = 0 \)
(d) \( x \pm y + 2 = 0 \)
Solution:
Tangent to the parabola
\( y = mx + \frac{2}{m} \) ...(1)
Tangent to the Hyp.
\( y = mx \pm \sqrt{m^2 - 3} \) ...(2)
(1) and (2) are same \( \frac{2}{m} = \pm\sqrt{m^2 - 3} \)
\( \implies m^4 - 3m^2 - 4 = 0 \)
\( \implies m^2 = 4 \)
\( \implies m = \pm 2 \)
From (1) \( 2x \pm y + 1 = 0 \)
Answer: (a) \( 2x \pm y + 1 = 0 \)

Question. Equation of the chord of the hyperbola \( 25x^2 - 16y^2 = 400 \) which is bisected at the point (6, 2) is
(a) \( 16x - 75y = 418 \)
(b) \( 75x - 16y = 418 \)
(c) \( 25x - 4y = 400 \)
(d) None of the options
Solution:
Slope of the chord \( = \frac{25}{16} \times \frac{x_1}{y_1} \)
\( = \frac{25}{16} \times \frac{6}{2} = \frac{75}{16} \)
Equation of chord passing through (6, 2)
\( y - 2 = \frac{75}{16}(x - 6) \)
\( 16y - 32 = 75x - 450 \)
\( 75x - 16y = 418 \)
Answer: (b) \( 75x - 16y = 418 \)

Question. Variable circles are drawn touching two fixed circles externally then locus of centre of variable circle is
(a) parabola
(b) ellipse
(c) hyperbola
(d) circle
Answer: (c) hyperbola
Solution:
CA - r_1 = r
CB - r_2 = r
CA - CB = r_1 - r_2 = k
CA - CB = k

\( \implies \) Locus of C will be hyperbola.

Question. The locus of the mid points of the chords passing through a fixed point \((\alpha, \beta)\) of the hyperbola, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is
(a) a circle with centre \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\)
(b) an ellipse with centre \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\)
(c) a hyperbola with centre \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\)
(d) straight line through \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\)
Answer: (c) a hyperbola with centre \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\)
Solution:
Let M (h, k)
Chord with given mid point (h, k)
\( T = S_1 \)
\( \implies \frac{hx}{a^2} - \frac{ky}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2} \)
\( (\alpha, \beta) \)
\( \implies \frac{h\alpha}{a^2} - \frac{k\beta}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2} \)
\( \frac{x\alpha}{a^2} - \frac{y\alpha}{b^2} = \frac{x^2}{a^2} - \frac{y^2}{b^2} \)
\( \frac{x^2}{a^2} - \frac{x\alpha}{a^2} - \left(\frac{y^2}{b^2} - \frac{y\beta}{b^2}\right) = 0 \)
\( \frac{x^2}{a^2} - \frac{x\alpha}{a^2} + \frac{\alpha^2}{4a^2} - \frac{\alpha^2}{4a^2} - \left(\frac{y^2}{b^2} - \frac{y\beta}{b^2} + \frac{\beta^2}{4b^2} - \frac{\beta^2}{4b^2}\right) = 0 \)
\( \left(\frac{x}{a} - \frac{\alpha}{2a}\right)^2 - \left(\frac{y}{b} - \frac{\beta}{2b}\right)^2 = \frac{\alpha^2}{4a^2} - \frac{\beta^2}{4b^2} \)
Centre will be \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\) And Hyperbola

Question. The locus of the foot of the perpendicular from the centre of the hyperbola \(xy = c^2\) on a variable tangent is
(a) \((x^2 - y^2)^2 = 4c^2xy\)
(b) \((x^2 + y^2)^2 = 2c^2xy\)
(c) \((x^2 - y^2) = 4c^2xy\)
(d) \((x^2 + y^2)^2 = 4c^2xy\)
Answer: (d) \((x^2 + y^2)^2 = 4c^2xy\)
Solution:
\( xy = c^2 \)
\( P\left(ct, \frac{c}{t}\right) \)
Tangent at P
\( \frac{x}{t} + ty = 2c \) ....(1)
Let N (h, k)
Slope of ON
\( m = t^2 = \frac{k}{h} \) ....(2)
point (h, k) lie on Equation (1)
\( \frac{h}{t} + tk = 2c \)

\( \implies (h + t^2k)^2 = 4c^2t^2 \)
\( \implies \left(h + \frac{k^2}{h}\right)^2 = 4c^2 \frac{k}{h} \)

\( \implies (h^2 + k^2)^2 = 4c^2 kh \)
\( \implies (x^2 + y^2)^2 = 4c^2 xy \)

Question. The equation to the chord joining two points \((x_1, y_1)\) and \((x_2, y_2)\) on the rectangular hyperbola \(xy = c^2\) is
(a) \(\frac{x}{x_1 + x_2} + \frac{y}{y_1 + y_2} = 1\)
(b) \(\frac{x}{x_1 - x_2} + \frac{y}{y_1 - y_2} = 1\)
(c) \(\frac{x}{y_1 - y_2} + \frac{y}{x_1 - x_2} = 1\)
(d) \(\frac{x}{y_1 - y_2} + \frac{y}{x_1 - x_2} = 1\)
Answer: (a) \(\frac{x}{x_1 + x_2} + \frac{y}{y_1 + y_2} = 1\)
Solution:
Let \( P(ct_1, c/t_1) \) \( Q(ct_2, c/t_2) \)
\( M_{PQ} = \frac{\frac{c}{t_2} - \frac{c}{t_1}}{c(t_2 - t_1)} = \frac{-1}{t_1t_2} \)
Equation \( y - \frac{c}{t_1} = \frac{-1}{t_1t_2} (x - ct_1) \)
\( x + t_1t_2y = c(t_1 + t_2) \)
\( \implies \frac{x}{c(t_1 + t_2)} + \frac{y}{c\left(\frac{1}{t_2} + \frac{1}{t_1}\right)} = 1 \)
\( \frac{x}{x_1 + x_2} + \frac{y}{y_1 + y_2} = 1 \)

Question. The equation \(9x^2 - 16y^2 - 18x + 32y - 151 = 0\) represent a hyperbola
(a) The length of the transverse axes is 4
(b) Length of latus rectum is 9
(c) Equation of directrix is \(x = \frac{21}{5}\) and \(x = -\frac{11}{5}\)
(d) None of the options
Answer: (c) Equation of directrix is \(x = \frac{21}{5}\) and \(x = -\frac{11}{5}\)
Solution:
\( 9x^2 - 16y^2 - 18x + 32y - 151 = 0 \)
\( 9(x^2 - 2x) - 16 (y^2 - 2y) - 151 = 0 \)
\( 9(x^2 - 2x + 1) - 9 - 16(y^2 - 2y + 1) + 16 - 151 = 0 \)
\( 9(x - 1)^2 - 16 (y - 1)^2 = 144 \)
\( \frac{(x - 1)^2}{\left(\frac{144}{9}\right)} - \frac{(y - 1)^2}{\left(\frac{144}{16}\right)} = 1 \)
\( \implies \frac{(x - 1)^2}{16} - \frac{(y - 1)^2}{9} = 1 \)
\( \ell(\text{TA}) = 2a = 8 \)
\( e^2 = 1 + \frac{b^2}{a^2} \)
\( \implies e = \frac{5}{4} \)
\( \ell(\text{LR}) = \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2} \)
Directries \( x - 1 = \frac{4}{\left(\frac{5}{4}\right)} \) and \( x - 1 = -\frac{16}{5} \)
\( x = \frac{21}{5} \)
\( x = -\frac{11}{5} \)

Question. From the points of the circle \(x^2 + y^2 = a^2\), tangents are drawn to the hyperbola \(x^2 - y^2 = a^2\); then the locus of the middle points of the chords of contact is
(a) \((x^2 - y^2)^2 = a^2(x^2 + y^2)\)
(b) \((x^2 - y^2)^2 = 2a^2(x^2 + y^2)\)
(c) \((x^2 + y^2)^2 = a^2(x^2 - y^2)\)
(d) \(2(x^2 - y^2)^2 = 3a^2(x^2 + y^2)\)
Answer: (a) \((x^2 - y^2)^2 = a^2(x^2 + y^2)\
Solution:
Let P(a cos\theta, a sin\theta)
Equation of QR (c.o.c. w.r.t. p) T = 0
\( x \cos \theta - y \sin \theta = a \) ...(1)
and T = S_1
\( hx - ky = h^2 - k^2 \) ...(2)
(1) and (2) are same
\( \frac{\cos\theta}{h} = \frac{\sin\theta}{k} = \frac{a}{h^2 - k^2} \)
square & add
\( (x^2 - y^2)^2 = a^2 (x^2 + y^2) \)

Question. The tangent to the hyperbola \(xy = c^2\) at the point P intersects the x-axis at T and the y-axis at T'. The normal to the hyperbola at P intersects the x-axis at N and the y-axis at N'. The areas of the triangles PNT and PN'T' are \(\Delta\) and \(\Delta'\) respectively, then \(\frac{1}{\Delta} + \frac{1}{\Delta'}\) is
(a) equal to 1
(b) depends on t
(c) depends on c
(d) equal to 2
Answer: (c) depends on c
Solution:
Tangent at P
\( \frac{x}{t} + ty = 2c \) ....(1)
Normal at P
\( y - \frac{c}{t} = xt^2 - ct^3 \) ....(2)
\( T (2ct, 0) \) ; \( T' (0, 2c/t) \)
\( N\left(ct - \frac{c}{t^3}, 0\right) \) ; \( N'\left(0, \frac{c}{t} - ct^3\right) \)
\( \Delta = \text{Area of } \Delta PNT = \frac{1}{2} \times \frac{c}{t} \left[2ct - ct + \frac{c}{t^3}\right] \)
\( \Delta = \frac{c^2}{2t^4} (t^4 + 1) \)
\( \Delta' = \text{Area of } \Delta PN'T' = \frac{1}{2} \times ct \times \left[\frac{2c}{t} - \frac{c}{t} + ct^3\right] = \frac{1}{2} c^2 (t^4 + 1) \)
\( \frac{1}{\Delta} + \frac{1}{\Delta'} = \frac{2}{c^2} \)

Question. The tangent to the hyperbola, \(x^2 - 3y^2 = 3\) at the point \((\sqrt{3}, 0)\) when associated with two asymptotes constitutes.
(a) isosceles triangle
(b) an equilateral triangle
(c) a triangle whose area is \(\sqrt{3}\) sq. unit
(d) a right isosceles triangle
Answer: (b) an equilateral triangle, (c) a triangle whose area is \(\sqrt{3}\) sq. unit
Solution:
\( \frac{x^2}{3} - \frac{y^2}{1} = 1 \)
Asyp \( y = \pm \frac{1}{\sqrt{3}} x \)
\( \Delta OPQ \) will be equilateral triangle.
PR = 1
area of \( \Delta OPQ = \frac{1}{2} \times \sqrt{3} \times (2) = \sqrt{3} \) sq. units

Question. The asymptote of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) form with any tangent to the hyperbola a triangle whose area is \(a^2 \tan \lambda\) in magnitude then its eccentricity is
(a) \(\sec \lambda\)
(b) \(\csc \lambda\)
(c) \(\sec^2 \lambda\)
(d) \(\csc^2 \lambda\)
Answer: (a) \(\sec \lambda\)
Solution:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
Let the any point be \((a, 0)\)
P(a, b), Q (a, -b)
PQ = 2b
OA = a
Area of \( \Delta OPA = \frac{1}{2} \times a \times 2b = ab \)
\( \implies ab = a^2 \tan\lambda \)
\( \implies \frac{b}{a} = \tan\lambda \)
\( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \tan^2\lambda} = \sec\lambda \)

Question. From any point on the hyperbola \(H_1 : (x^2/a^2) - (y^2/b^2) = 1\) tangents are drawn to the hyperbola \(H_2 : (x^2/a^2) - (y^2/b^2) = 2\). The area cut-off by the chord of contact on the asymptotes of \(H_2\) is equal to
(a) \(ab/2\)
(b) \(ab\)
(c) \(2 ab\)
(d) \(4 ab\)
Answer: (d) \(4 ab\)
Solution:
Let the point \((a \sec \theta, b \tan \theta)\)
C.O.C. : \( \frac{x}{a} \sec\theta - \frac{y}{b} \tan\theta = 2 \) ....(1)
PoI of asymptotes and Eq^n (1)
\( A[2a (\sec\theta + \tan\theta), 2b (\sec\theta + \tan\theta)] \)
\( B[2a (\sec\theta - \tan\theta), -2b (\sec\theta - \tan\theta)] \)
Area of Triangle OAB = \( \frac{1}{2} (8ab) = 4ab \)

Question. The tangent at P on the hyperbola \((x^2/a^2) - (y^2/b^2) = 1\) meets the asymptote \(\frac{x}{a} - \frac{y}{b} = 0\) at Q. If the locus of the mid point of PQ has the equation \((x^2/a^2) - (y^2/b^2) = k\), then k has the value equal to
(a) 1/2
(b) 2
(c) 3/4
(d) 4/3
Answer: (c) 3/4
Solution:
\( P(a, 0) \) ; \( Q(a, b) \)
Let M (h, k)
\( 2h = 2a \)
\( \implies h = a \)
\( k = \frac{b}{2} \)
\( \left(\frac{h}{a}\right)^2 - \left(\frac{k}{b}\right)^2 = 1 - \frac{1}{4} \)
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{3}{4} \) So \( k = \frac{3}{4} \)

Question. If \(\theta\) is the angle between the asymptotes of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) with eccentricity e, then \(\sec \theta/2\) can be
(a) e
(b) e/2
(c) e/3
(d) 1/e
Answer: (a) e, (d) 1/e
Solution:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) Asyp. \( y = \pm \frac{b}{a} x \)
\( m_1 = \frac{b}{a} \) and \( m_2 = -\frac{b}{a} \)
\( \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{\frac{b}{a} + \frac{b}{a}}{1 - \frac{b^2}{a^2}}\right| \)
\( \tan\theta = \frac{2ab}{a^2 - b^2} \)
\( \implies \tan\frac{\theta}{2} = \frac{b}{a} \) and \( -\frac{a}{b} \)
\( \sec\frac{\theta}{2} = \sqrt{1 + \frac{b^2}{a^2}} \) and \( \sec\frac{\theta}{2} = \sqrt{1 + \frac{a^2}{b^2}} \)
= e = \( \frac{1}{e} \)

Question. If (5, 12) and (24, 7) are the focii of a conic passing through the origin then the eccentricity of conic is
(a) \(\sqrt{386}/12\)
(b) \(\sqrt{386}/13\)
(c) \(\sqrt{386}/25\)
(d) \(\sqrt{386}/38\)
Answer: (a) \(\sqrt{386}/12\), (d) \(\sqrt{386}/38\)
Solution:
mid pt. of \( F_1F_2 \)
\( \implies \) Centre of conic
\( C \left(\frac{29}{2}, \frac{19}{2}\right) \)
equation \( \frac{\left(x - \frac{29}{2}\right)^2}{a^2} - \frac{\left(y - \frac{19}{2}\right)^2}{b^2} = 1 \)
passing through origin
\( \frac{(29)^2}{4a^2} - \frac{(19)^2}{4b^2} = 1 \)
\( 2ae = \sqrt{386} \)
Solve & get e

Question. The point of contact of \(5x + 12y = 19\) and \(x^2 - 9y^2 = 9\) will lie on
(a) \(4x + 15y = 0\)
(b) \(7x + 12y = 19\)
(c) \(4x + 15y + 1 = 0\)
(d) \(7x - 12y = 19\)
Answer: (a) \(4x + 15y = 0\), (b) \(7x + 12y = 19\)
Solution:
Let the point \( P(x_1, y_1) \)
tangent at P
\( xx_1 - 9yy_1 = 9 \)
\( x \left(\frac{x_1}{9}\right) - y (y_1) = 1 \) ....(1)
\( \left(\frac{5}{19}\right) x + \left(\frac{12}{19}\right) y = 1 \) ....(2)
By comparing (1) & (2)
\( x_1 = \frac{45}{19} : y_1 = \frac{-12}{19} \)

Question. Equation \((2 + \lambda)x^2 - 2\lambda xy + (\lambda - 1)y^2 - 4x - 2 = 0\) represents a hyperbola if
(a) \(\lambda = 4\)
(b) \(\lambda = 1\)
(c) \(\lambda = 4/3\)
(d) \(\lambda = -1\)
Answer: (b) \(\lambda = 1\), (d) \(\lambda = -1\)
Solution:
Hyperbola if
\( h^2 > ab \)
\( \implies \lambda^2 > (2 + \lambda) (\lambda - 1) \)
\( \implies \lambda < 2 \)
and \( D \ne 0 \)
\( \implies -2 [3\lambda - 4] \ne 0 \)
\( \implies \lambda \ne 4/3 \)

Question. The curve described parametrically by, \( x=t^2 +t+1, y = t^2-t+1 \) represents 
(a) a parabola
(b) an ellipse
(c) a hyperbola
(d) a pair of straight lines
Answer: (a) a parabola
Solution:
\( x = t^2 + t + 1 \)
\( y = t^2 - t + 1 \) ; \( \frac{x - y}{2} = t \) ; \( \frac{x + y}{2} = t^2 + 1 \)
Eliminate t
\( 2(x + y) = (x - y)^2 + 4 \)
It is a perfect square. So it is a parabola (A)

Question. Let P \( (a \sec \theta, b \tan \theta) \) and Q \( (a \sec \phi, b \tan \phi) \), where \( \theta + \phi = \frac{\pi}{2} \), be two points on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). If \( (h, k) \) is the point of intersection of the normals at P & Q, then k is equal to
(a) \( \frac{a^2 + b^2}{a} \)
(b) \( -\left( \frac{a^2 + b^2}{a} \right) \)
(c) \( \frac{a^2 + b^2}{b} \)
(d) \( -\left( \frac{a^2 + b^2}{b} \right) \)
Answer: (d) \( -\left( \frac{a^2 + b^2}{b} \right) \)
Solution:
\( \phi = \frac{\pi}{2} - \theta \)
Normal at \( \theta, \phi \) passes through \( (h, k) \)
\( \therefore ah \cos\theta + bk \cot \theta = a^2 + b^2 \)
\( ah \sin \theta + bk \tan \theta = a^2 + b^2 \)
Eliminate h
\( bk (\cot \theta \sin \theta - \tan \theta \cos \theta) = (a^2 + b^2) (\sin \theta - \cos\theta) \)
so \( k = - \frac{(a^2 + b^2)}{b} \)

Question. If x= 9 is the chord of contact of the hyperbola \( x^2 - y^2 = 9 \), then the equation of the corresponding pair of tangents, is
(a) \( 9x^2 - 8y^2 + 18x - 9 = 0 \)
(b) \( 9x^2 - 8y^2 - 18x + 9 = 0 \)
(c) \( 9x^2 - 8y^2 - 18x - 9 = 0 \)
(d) \( 9x^2 - 8y^2 + 18x + 9 = 0 \)
Answer: (c) \( 9x^2 - 8y^2 - 18x - 9 = 0 \)
Solution:
c.o.c. with respect to \( (h, k) \)
\( hx - ky = 9 \)
compare with x = 9
h = 1, k = 0
(1, 0)
pair of Tangents
\( SS_1 = T^2 \)
\( (x^2 - y^2 - 9) (1 + 0 - 9) = (x - 9)^2 \)
\( -8x^2 + 8y^2 + 72 = x^2 - 18x + 81 \)
\( 9x^2 - 8y^2 - 18x + 9 = 0 \)

Question. The equation of the common tangent to the curve \( y^2 = 8x \) and \( xy = -1 \) is
(a) \( 3y = 9x + 2 \)
(b) \( y = 2x + 1 \)
(c) \( 2y = x + 8 \)
(d) \( y = x + 2 \)
Answer: (d) \( y = x + 2 \)
Solution:
Tangent to the parabola \( y^2 = 8x \)
\( y = mx + \frac{2}{m} \) ...(1)
It will also touch \( xy = -1 \)
\( x \left( mx + \frac{2}{m} \right) = -1 \)
Has equal roots
\( \frac{4}{m^2} = 4m \)
\( \implies \) \( m = 1 \)
So \( y = x + 2 \)

Question. Given the family of hyperbolas \( \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1 \) for \( \alpha \in (0, \pi/2) \) which of the following does not change with varying \( \alpha \) ?
(a) abscissa of foci
(b) eccentricity
(c) equations of directrices
(d) abscissa of vertices
Answer: (a) abscissa of foci
Solution:
\( \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1 \) for \( \alpha \in \left( 0, \frac{\pi}{2} \right) \)
\( \sin^2 \alpha = \cos^2\alpha (e^2 - 1) \)
\( \cos^2\alpha e^2 = 1 \)
abscissa of foci \( (\cos\alpha e, 0) \) does not change.

Question. The line \( 2x + \sqrt{6} y= 2 \) is a tangent to the curve \( x^2 - 2y^2 = 4 \). The point of contact is
(a) \( (4, -\sqrt{6}) \)
(b) \( (7, -2\sqrt{6}) \)
(c) \( (2, 3) \)
(d) \( (\sqrt{6}, 1) \)
Answer: (a) \( (4, -\sqrt{6}) \)
Solution:
\( 2x + \sqrt{6} y = 2 \) ...(1)
Hyp. \( x^2 - 2y^2 = 4 \)
Equation of tangent
\( xx_1 - 2yy_1 = 4 \)
Comparing with (1)
\( x_1 = 4 \) and \( y_1 = -\sqrt{6} \)