Question

A hyperbola having the transverse axis of the length \[2\sin\theta\] is confocal (foci of both conic coincide) with the ellipse \[3x^{2}+4y^{2}=12.\] Then, its equation is______.

Options

• \[x^{2}cosec^{2}\theta-y^{2}\sec^{2}\theta=1\]

• \[x^{2}\sec^{2}\theta-y^{2}cosec^{2}\theta=1\]

• \[x^{2}\sin^{2}\theta-y^{2}\cos^{2}\theta=1\]

• \[x^{2}\sin^{2}\theta-y^{2}\sin^{2}\theta=1\]

Answer 1

A hyperbola having the transverse axis of the length \[2\sin\theta\] is confocal (foci of both conic coincide) with the ellipse \[3x^{2}+4y^{2}=12.\] Then, its equation is \[x^{2}cosec^{2}\theta-y^{2}\sec^{2}\theta=1\].

Explanation:

Given, \[3x^{2}+4y^{2}=12\]

\[\Rightarrow\quad\frac{x^2}{4}+\frac{y^2}{3}=1\]

\[\therefore\quad a^2=4\Rightarrow a=2\mathrm{and}b^2=3\Longrightarrow b=\sqrt{3}\]

\[\therefore\quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\]

\[\because\quad ae=2\times\frac{1}{2}=1\]

Given, \[a=\sin\theta\]

\[\Rightarrow(\sin\theta)e=1\]

\[\Rightarrow\quad e=\mathrm{cosec}\theta\]

and \[b=a\sqrt{e^2-1}\]

         \[=\sin\theta\sqrt{\mathrm{cosec}^2\theta-1}\]

        \[=\sin\theta\times\cot\theta=\cos\theta\]

\[\therefore\] Equation of hyperbola is \[\frac{x^2}{\sin^2\theta}-\frac{y^2}{\cos^2\theta}=1\]

\[\Rightarrow\quad x^{2}\mathrm{cosec}^{2}\theta-y^{2}\sec^{2}\theta=1\]