\displaystyle \textbf{Question 1. }\text{Equation of the hyperbola whose vertices are }(\pm3,0)\text{ and foci at }(\pm5,0),\text{ is}
\displaystyle \text{(a) }16x^2-9y^2=144\qquad \text{(b) }9x^2-16y^2=144\qquad  \\ \text{(c) }25x^2-9y^2=225\qquad \text{(d) }9x^2-25y^2=81
\displaystyle \text{Answer:}
\displaystyle a=3,\quad c=5
\displaystyle c^2=a^2+b^2
\displaystyle 25=9+b^2
\displaystyle b^2=16
\displaystyle \therefore \frac{x^2}{9}-\frac{y^2}{16}=1
\displaystyle \Rightarrow 16x^2-9y^2=144
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 2. }\text{If }e_1\text{ and }e_2\text{ are respectively the eccentricities of the ellipse }
\displaystyle \frac{x^2}{18}+\frac{y^2}{4}=1  \text{and the hyperbola }\frac{x^2}{9}-\frac{y^2}{4}=1,\text{ then the relation between }e_1\text{ and }e_2\text{ is}
\displaystyle \text{(a) }3e_1^2+e_2^2=2\qquad \text{(b) }e_1^2+2e_2^2=3\qquad \text{(c) }2e_1^2+e_2^2=3\qquad \text{(d) }e_1^2+3e_2^2=2
\displaystyle \text{Answer:}
\displaystyle \text{For ellipse, }a^2=18,\quad b^2=4
\displaystyle e_1^2=1-\frac{b^2}{a^2}=1-\frac{4}{18}=\frac{7}{9}
\displaystyle \text{For hyperbola, }a^2=9,\quad b^2=4
\displaystyle e_2^2=1+\frac{b^2}{a^2}=1+\frac{4}{9}=\frac{13}{9}
\displaystyle 2e_1^2+e_2^2=2\left(\frac{7}{9}\right)+\frac{13}{9}
\displaystyle =\frac{27}{9}=3
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 3. }\text{The distance between the directrices of the hyperbola } \\ x=8\sec\theta,\ y=8\tan\theta,\text{ is}
\displaystyle \text{(a) }8\sqrt2\qquad \text{(b) }16\sqrt2\qquad \text{(c) }4\sqrt2\qquad \text{(d) }6\sqrt2
\displaystyle \text{Answer:}
\displaystyle x=8\sec\theta,\quad y=8\tan\theta
\displaystyle \frac{x^2}{64}-\frac{y^2}{64}=\sec^2\theta-\tan^2\theta=1
\displaystyle a^2=64,\quad b^2=64
\displaystyle a=8,\quad e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt2
\displaystyle \text{Directrices are }x=\pm\frac{a}{e}
\displaystyle \text{Distance between directrices}=\frac{2a}{e}
\displaystyle =\frac{16}{\sqrt2}=8\sqrt2
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 4. }\text{The equation of the conic with focus at }(1,-1),\text{ directrix along } \\ x-y+1=0\text{ and eccentricity }\sqrt2\text{ is}
\displaystyle \text{(a) }xy=1\qquad \text{(b) }2xy+4x-4y-1=0\qquad \text{(c) }x^2-y^2=1\qquad \text{(d) }2xy-4x+4y+1=0
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be any point on the conic.}
\displaystyle \text{Then }SP=e\cdot PM
\displaystyle \sqrt{(x-1)^2+(y+1)^2}=\sqrt2\cdot\frac{|x-y+1|}{\sqrt{1^2+(-1)^2}}
\displaystyle \sqrt{(x-1)^2+(y+1)^2}=|x-y+1|
\displaystyle (x-1)^2+(y+1)^2=(x-y+1)^2
\displaystyle x^2-2x+1+y^2+2y+1=x^2+y^2+1-2xy+2x-2y
\displaystyle 2xy-4x+4y+1=0
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 5. }\text{The eccentricity of the conic }9x^2-16y^2=144\text{ is}
\displaystyle \text{(a) }\frac{5}{4}\qquad \text{(b) }\frac{4}{3}\qquad \text{(c) }\frac{4}{5}\qquad \text{(d) }\sqrt7
\displaystyle \text{Answer:}
\displaystyle 9x^2-16y^2=144
\displaystyle \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1
\displaystyle a^2=16,\quad b^2=9
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}
\displaystyle =\sqrt{1+\frac{9}{16}}
\displaystyle =\sqrt{\frac{25}{16}}=\frac{5}{4}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 6. }\text{A point moves in a plane so that its distances }PA\text{ and }PB\text{ from} \\ \text{two fixed points }A\text{ and }B  \text{in the plane satisfy the relation }PA-PB=k\ (k\neq0),\text{ then the} \\ \text{locus of }P\text{ is}
\displaystyle \text{(a) a hyperbola}\qquad \text{(b) a branch of the hyperbola}\qquad \text{(c) a parabola}\qquad \text{(d) an ellipse}
\displaystyle \text{Answer:}
\displaystyle \text{The difference of distances of a moving point from two fixed points is constant.}
\displaystyle \text{This is the defining property of a hyperbola.}
\displaystyle \text{Since }PA-PB=k\text{ has a fixed sign, it represents one branch of the hyperbola.}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 7. }\text{The eccentricity of the hyperbola whose latus-rectum is half of its} \\ \text{transverse axis, is}
\displaystyle \text{(a) }\frac{1}{\sqrt2}\qquad \text{(b) }\sqrt{\frac{2}{3}}\qquad \text{(c) }\sqrt{\frac{3}{2}}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}
\displaystyle \text{Length of transverse axis}=2a
\displaystyle \frac{2b^2}{a}=\frac{1}{2}(2a)
\displaystyle \frac{2b^2}{a}=a
\displaystyle 2b^2=a^2
\displaystyle \frac{b^2}{a^2}=\frac{1}{2}
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}
\displaystyle =\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 8. }\text{The eccentricity of the hyperbola }x^2-4y^2=1\text{ is}
\displaystyle \text{(a) }\frac{\sqrt3}{2}\qquad \text{(b) }\frac{\sqrt5}{2}\qquad \text{(c) }\frac{2}{\sqrt3}\qquad \text{(d) }\frac{2}{\sqrt5}
\displaystyle \text{Answer:}
\displaystyle x^2-4y^2=1
\displaystyle \Rightarrow \frac{x^2}{1}-\frac{y^2}{\frac{1}{4}}=1
\displaystyle a^2=1,\quad b^2=\frac{1}{4}
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}
\displaystyle =\sqrt{1+\frac{1}{4}}
\displaystyle =\frac{\sqrt5}{2}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 9. }\text{The difference of the focal distances of any point on the hyperbola} \\ \text{is equal to}
\displaystyle \text{(a) length of the conjugate axis}\qquad \text{(b) eccentricity}
\displaystyle \text{(c) length of the transverse axis}\qquad \text{(d) Latus-rectum}
\displaystyle \text{Answer:}
\displaystyle \text{For a hyperbola, the absolute difference of the focal distances of any point is constant.}
\displaystyle \text{This constant is }2a
\displaystyle \therefore \text{It is equal to the length of the transverse axis.}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 10. }\text{The foci of the hyperbola }9x^2-16y^2=144\text{ are}
\displaystyle \text{(a) }(\pm4,0)\qquad \text{(b) }(0,\pm4)\qquad \text{(c) }(\pm5,0)\qquad \text{(d) }(0,\pm5)
\displaystyle \text{Answer:}
\displaystyle 9x^2-16y^2=144
\displaystyle \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1
\displaystyle a^2=16,\quad b^2=9
\displaystyle c^2=a^2+b^2=16+9=25
\displaystyle c=5
\displaystyle \therefore \text{Foci}=(\pm5,0)
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 11. }\text{The distance between the foci of a hyperbola is }16\text{ and its eccentricity is }\sqrt2,\text{ then equation of the hyperbola is}
\displaystyle \text{(a) }x^2+y^2=32\qquad \text{(b) }x^2-y^2=16\qquad \text{(c) }x^2+y^2=16\qquad \text{(d) }x^2-y^2=32
\displaystyle \text{Answer:}
\displaystyle \text{Distance between foci}=2ae=16
\displaystyle ae=8
\displaystyle e=\sqrt2
\displaystyle a=\frac{8}{\sqrt2}=4\sqrt2
\displaystyle a^2=32
\displaystyle e^2=1+\frac{b^2}{a^2}
\displaystyle 2=1+\frac{b^2}{32}
\displaystyle b^2=32
\displaystyle \therefore \frac{x^2}{32}-\frac{y^2}{32}=1
\displaystyle \Rightarrow x^2-y^2=32
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 12. }\text{If }e_1\text{ is the eccentricity of the conic }9x^2+4y^2=36\text{ and }e_2\text{ is} \\ \text{the eccentricity of the conic }9x^2-4y^2=36,
\displaystyle \text{then}
\displaystyle \text{(a) }e_1^2-e_2^2=2\qquad \text{(b) }2<e_2^2-e_1^2<3\qquad \text{(c) }e_2^2-e_1^2=2\qquad \text{(d) }e_2^2-e_1^2>3
\displaystyle \text{Answer:}
\displaystyle 9x^2+4y^2=36
\displaystyle \Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1
\displaystyle \text{For ellipse, }a^2=9,\quad b^2=4
\displaystyle e_1^2=1-\frac{b^2}{a^2}=1-\frac{4}{9}=\frac{5}{9}
\displaystyle 9x^2-4y^2=36
\displaystyle \Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1
\displaystyle \text{For hyperbola, }a^2=4,\quad b^2=9
\displaystyle e_2^2=1+\frac{b^2}{a^2}=1+\frac{9}{4}=\frac{13}{4}
\displaystyle e_2^2-e_1^2=\frac{13}{4}-\frac{5}{9}
\displaystyle =\frac{117-20}{36}=\frac{97}{36}
\displaystyle 2<\frac{97}{36}<3
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 13. }\text{If the eccentricity of the hyperbola }x^2-y^2\sec^2\alpha=5\text{ is } \\ \sqrt3\text{ times the eccentricity of the ellipse}
\displaystyle x^2\sec^2\alpha+y^2=25,\text{ then }\alpha=
\displaystyle \text{(a) }\frac{\pi}{6}\qquad \text{(b) }\frac{\pi}{4}\qquad \text{(c) }\frac{\pi}{3}\qquad \text{(d) }\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle x^2-y^2\sec^2\alpha=5
\displaystyle \Rightarrow \frac{x^2}{5}-\frac{y^2}{5\cos^2\alpha}=1
\displaystyle e_1=\sqrt{1+\frac{5\cos^2\alpha}{5}}=\sqrt{1+\cos^2\alpha}
\displaystyle x^2\sec^2\alpha+y^2=25
\displaystyle \Rightarrow \frac{x^2}{25\cos^2\alpha}+\frac{y^2}{25}=1
\displaystyle e_2=\sqrt{1-\frac{25\cos^2\alpha}{25}}=\sqrt{1-\cos^2\alpha}=\sin\alpha
\displaystyle \text{Given }e_1=\sqrt3e_2
\displaystyle \sqrt{1+\cos^2\alpha}=\sqrt3\sin\alpha
\displaystyle 1+\cos^2\alpha=3\sin^2\alpha
\displaystyle 1+\cos^2\alpha=3(1-\cos^2\alpha)
\displaystyle 4\cos^2\alpha=2
\displaystyle \cos^2\alpha=\frac{1}{2}
\displaystyle \therefore \alpha=\frac{\pi}{4}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 14. }\text{The equation of the hyperbola whose foci are }(6,4)\text{ and }(-4,4) \\ \text{ and eccentricity }2,\text{ is}
\displaystyle \text{(a) }\frac{(x-1)^2}{25/4}-\frac{(y-4)^2}{75/4}=1\qquad \text{(b) }\frac{(x+1)^2}{25/4}-\frac{(y+4)^2}{75/4}=1
\displaystyle \text{(c) }\frac{(x-1)^2}{75/4}-\frac{(y-4)^2}{25/4}=1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Centre is the midpoint of the foci}
\displaystyle C=\left(\frac{6-4}{2},\frac{4+4}{2}\right)=(1,4)
\displaystyle \text{Distance between foci}=10
\displaystyle \therefore 2ae=10
\displaystyle ae=5
\displaystyle e=2
\displaystyle \therefore a=\frac{5}{2}
\displaystyle a^2=\frac{25}{4}
\displaystyle c^2=a^2+b^2,\quad c=ae=5
\displaystyle 25=\frac{25}{4}+b^2
\displaystyle b^2=\frac{75}{4}
\displaystyle \therefore \frac{(x-1)^2}{25/4}-\frac{(y-4)^2}{75/4}=1
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 15. }\text{The length of the straight line }x-3y=1\text{ intercepted by the} \\ \text{hyperbola }x^2-4y^2=1\text{ is}
\displaystyle \text{(a) }\frac{6}{\sqrt5}\qquad \text{(b) }3\sqrt{\frac{2}{5}}\qquad \text{(c) }6\sqrt{\frac{2}{5}}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x-3y=1
\displaystyle \Rightarrow x=1+3y
\displaystyle \text{Substituting in }x^2-4y^2=1,
\displaystyle (1+3y)^2-4y^2=1
\displaystyle 1+6y+9y^2-4y^2=1
\displaystyle 5y^2+6y=0
\displaystyle y(5y+6)=0
\displaystyle y=0\quad \text{or}\quad y=-\frac{6}{5}
\displaystyle \text{Corresponding }x=1\quad \text{or}\quad x=-\frac{13}{5}
\displaystyle \text{Required length}=\sqrt{\left(1+\frac{13}{5}\right)^2+\left(0+\frac{6}{5}\right)^2}
\displaystyle =\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{6}{5}\right)^2}
\displaystyle =\sqrt{\frac{360}{25}}=\frac{6\sqrt{10}}{5}=6\sqrt{\frac{2}{5}}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 16. }\text{The latus-rectum of the hyperbola }16x^2-9y^2=144\text{ is}
\displaystyle \text{(a) }\frac{16}{3}\qquad \text{(b) }\frac{32}{3}\qquad \text{(c) }\frac{8}{3}\qquad \text{(d) }\frac{4}{3}
\displaystyle \text{Answer:}
\displaystyle 16x^2-9y^2=144
\displaystyle \Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1
\displaystyle a^2=9,\quad b^2=16
\displaystyle a=3
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}
\displaystyle =\frac{2(16)}{3}=\frac{32}{3}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 17. }\text{The foci of the hyperbola }2x^2-3y^2=5\text{ are}
\displaystyle \text{(a) }\left(\pm\frac{5}{\sqrt6},0\right)\qquad \text{(b) }\left(\pm\frac{5}{6},0\right)\qquad \text{(c) }\left(\pm\frac{\sqrt5}{6},0\right)\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2x^2-3y^2=5
\displaystyle \Rightarrow \frac{x^2}{5/2}-\frac{y^2}{5/3}=1
\displaystyle a^2=\frac{5}{2},\quad b^2=\frac{5}{3}
\displaystyle c^2=a^2+b^2=\frac{5}{2}+\frac{5}{3}=\frac{25}{6}
\displaystyle c=\frac{5}{\sqrt6}
\displaystyle \therefore \text{Foci}=\left(\pm\frac{5}{\sqrt6},0\right)
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 18. }\text{The eccentricity of the hyperbola }x=\frac{a}{2}\left(t+\frac{1}{t}\right),\ y=\frac{a}{2}\left(t-\frac{1}{t}\right)\text{ is}
\displaystyle \text{(a) }\sqrt2\qquad \text{(b) }\sqrt3\qquad \text{(c) }2\sqrt3\qquad \text{(d) }3\sqrt2
\displaystyle \text{Answer:}
\displaystyle x+y=at,\quad x-y=\frac{a}{t}
\displaystyle \therefore (x+y)(x-y)=a^2
\displaystyle x^2-y^2=a^2
\displaystyle \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{a^2}=1
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+1}=\sqrt2
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 19. }\text{The equation of the hyperbola whose centre is }(6,2),\text{ one focus is } \\ (4,2)\text{ and eccentricity }2\text{ is}
\displaystyle \text{Answer:}
\displaystyle c=2,\quad e=2
\displaystyle ae=c
\displaystyle 2a=2\Rightarrow a=1
\displaystyle b^2=c^2-a^2=4-1=3
\displaystyle \therefore \frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1
\displaystyle \Rightarrow 3(x-6)^2-(y-2)^2=3
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 20. }\text{The locus of the point of intersection of the lines }
\displaystyle \sqrt3x-y-4\sqrt3\lambda=0  \text{ and }\sqrt3\lambda x+\lambda y-4\sqrt3=0\text{ is a hyperbola of eccentricity}
\displaystyle \text{Answer:}
\displaystyle \sqrt3x-y-4\sqrt3\lambda=0
\displaystyle \Rightarrow \lambda=\frac{\sqrt3x-y}{4\sqrt3}
\displaystyle \sqrt3\lambda x+\lambda y-4\sqrt3=0
\displaystyle \Rightarrow \lambda(\sqrt3x+y)=4\sqrt3
\displaystyle \frac{\sqrt3x-y}{4\sqrt3}(\sqrt3x+y)=4\sqrt3
\displaystyle 3x^2-y^2=48
\displaystyle \Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1
\displaystyle e=\sqrt{1+\frac{48}{16}}=\sqrt4=2
\displaystyle \therefore \text{Correct option is (b)}
\\

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