Question

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{{\alpha }^2}-\frac{y^2}{{\beta }^2}=1$ cut each other orthogonally, then

• A. $a^2+b^2={\alpha }^2+{\beta }^2$
• B. $a^2-b^2={\alpha }^2-{\beta }^2$
• C. $a^2-b^2={\alpha }^2+{\beta }^2$
• D. $a^2+b^2={\alpha }^2-{\beta }^2$

Answer 1

Answer: (C)

$a^2-b^2={\alpha }^2+{\beta }^2$

Slope of $I=\frac{-b^2{x}_1}{a^2{y}_1}=m_1\left(say\right)$
And slope of $II=\frac{{\beta }^2{x}_1}{{\alpha }^2{y}_1}=m_2\left(say\right)$
Using condition of orthogonality, $m_1{m}_2=-1$
$\frac{x_1^2}{y_1^2}=\frac{{\alpha }^2{a}^2}{{\beta }^2{b}^2}$
Also $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1$
and $\frac{x_1^2}{{\alpha }^2}-\frac{y_1^2}{{\beta }^2}=1$
$\Rightarrow x_1^2(\frac{1}{a^2}-\frac{1}{{\alpha }^2})=-y_1^2(\frac{1}{b^2}+\frac{1}{{\beta }^2})$
$\Rightarrow \frac{\frac{1}{a^2}-\frac{1}{{\alpha }^2}}{b^2{\beta }^2}=\frac{-(\frac{1}{b^2}+\frac{1}{{\beta }^2})}{{\alpha }^2{a}^2}$
$\therefore {\alpha }^2+{\beta }^2=a^2-b^2$
Hence, option 'C' is correct.