Question

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

• A. $a^2+b^2=p^2+m^2$
• B. $a^2-b^2=l^2-m^2$
• C. $a^2-b^2=p^2+m^2$
• D. $a^2+b^2=p^2-m^2$

Answer 1

The correct option is C $a^2-b^2=p^2+m^2$

For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$y^{{}^{\prime }}=-\frac{x{b}^2}{a^{2}y}=m_1$

For $\frac{x^2}{p^2}-\frac{y^2}{m^2}=1$
$y^{{}^{\prime }}=\frac{x{m}^2}{p^{2}y}=m_2$

$\because m_1{m}_2=-1$

$\frac{b^2{m}^2{x}^2}{p^2{a}^2{y}^2}=-1$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{x^2}{p^2}-\frac{y^2}{m^2}$

$\Rightarrow \frac{x^2}{a^2}-\frac{x^2}{p^2}=-(\frac{1}{b^2}+\frac{1}{m^2})y^2$

$\Rightarrow x^2(\frac{1}{p^2}-\frac{1}{a^2})=(\frac{1}{b^2}+\frac{1}{m^2})y^2$

$\Rightarrow p^2{a}^2(\frac{1}{p^2}-\frac{1}{a^2})=(\frac{1}{b^2}+\frac{1}{m^2})b^2{m}^2$

$\Rightarrow a^2-b^2=p^2+m^2$

Hence, option C.