Q97 / 160 · TS EAMCET 2013 · Mathematics
If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$, then the equation of the tangent at one of the points of contact is
- A$x+\sqrt{3} y+2=0$
- Correct answerB$3 x+2 \sqrt{2} y-3=0$
- C$3 x-\sqrt{2} y+6=0$
- D$x-\sqrt{3} y+2=0$
Solution
Given that, $x=9$ is a chord of contact of hyperbola.
$$
\begin{aligned}
& x^2-y^2=9 \\
& \text { put } x=9, \quad 81-y^2=9 \\
& \Rightarrow \quad y^2=72 \\
& \Rightarrow \quad y=6 \sqrt{2} \text { or }-6 \sqrt{2} \\
&
\end{aligned}
$$
$\therefore$ Points are $(9,6 \sqrt{2})$ and $(9,-6 \sqrt{2})$
Now, differentiating Eq. (i) w.r.t. $x$, we get
$$
\begin{aligned}
& 2 x-2 y \frac{d y}{d x}=0 \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{x}{y} \\
& \text { at }(9,6 \sqrt{2}) \quad\left(\frac{d y}{d x}\right)_{(9,6 \sqrt{2})}=\frac{9}{6 \sqrt{2}}=\frac{3}{2 \sqrt{2}} \\
& \text { and at }(9-6 \sqrt{2})\left(\frac{d y}{d x}\right)_{(9,-6 \sqrt{2})}=-\frac{3}{2 \sqrt{2}} \\
&
\end{aligned}
$$
$\therefore$ Equation of tangent at $(9,6 \sqrt{2})$ is
$$
\begin{aligned}
(y-6 \sqrt{2}) & =\frac{3}{2 \sqrt{2}}(x-9) \\
\Rightarrow \quad 2 \sqrt{2} y-24 & =3 x-27 \\
\Rightarrow \quad 3 x-2 \sqrt{2} y-3 & =0
\end{aligned}
$$
and equation of tangent at $(9,-6 \sqrt{2})$ is
$$
\begin{aligned}
(y+6 \sqrt{2}) & =\frac{-3}{2 \sqrt{2}}(x-9) \\
\Rightarrow \quad 2 \sqrt{2} y+24 & =-3 x+27 \\
\Rightarrow \quad 3 x+2 \sqrt{2} y-3 & =0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.