Q68 / 120 · VITEEE 2012 · Mathematics
If $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ and $x^{2}-y^{2}=c^{2}$ cut at right angles, then
- A$a^{2}+b^{2}=2 c^{2}$
- B$b^{2}-a^{2}=2 c^{2}$
- Correct answerC$a^{2}-b^{2}=2 c^{2}$
- D$a^{2} b^{2}=2 c^{2}$
Solution
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$..(i)
On differentiating w.r.t. $x$, we get
$$
\begin{array}{l}
\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0 \\
\Rightarrow \frac{d y}{d x}=-\frac{x b^{2}}{a^{2} y} \text { and } \\
x^{2}-y^{2}=c^{2}
\end{array}
$$
On differentiating w.r.t. $x$, we get
$$
\begin{array}{l}
2 x-2 y \frac{d y}{d x}=0 \\
\Rightarrow \frac{d y}{d x}=\frac{x}{y}
\end{array}
$$
The two curves will cut at right angles, if
$$
\begin{array}{l}
\left(\frac{d y}{d x}\right)_{c_{1}} \times\left(\frac{d y}{d x}\right)_{c_{2}}=-1 \\
\Rightarrow-\frac{b^{2} x}{a^{2} y} \cdot \frac{x}{y}=-1 \\
\Rightarrow \frac{x^{2}}{a^{2}}=\frac{y^{2}}{b^{2}} \\
\Rightarrow \frac{x^{2}}{a^{2}}=\frac{y^{2}}{b^{2}}=\frac{1}{2}
\end{array}
$$
[usingeq. (i)]
On substituting these values in $x^{2}-y^{2}=$
$\mathrm{c}^{2}$, we get
$$
\begin{array}{l}
\frac{a^{2}}{2}-\frac{b^{2}}{2}=c^{2} \\
\Rightarrow a^{2}-b^{2}=2 c^{2}
\end{array}
$$
Hence, the correct answer is option: 'C'.