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,

• 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

The correct option is A $a^2+b^2={\alpha }^2+{\beta }^2$

Slope of $I=\frac{-b^{2}x}{a^{2}y}=m_1$ (say),
And slope of II =$\frac{{\beta }^2}{{\alpha }^{2}y}=m_2\left(say\right)\therefore m_1{m}_2=1\therefore b^2{\beta }^2{x}^2=a^2{\alpha }^2{y}^2$
and $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$
$\frac{x^2}{{\alpha }^2}-\frac{y^2}{{\beta }^2}-1=0\Rightarrow x^2(\frac{1}{a^2}-\frac{1}{{\alpha }^2})=-y^2(\frac{1}{b^2}-\frac{1}{{\beta }^2})$
From Eqs. (i) and (ii)
$\Rightarrow \frac{\frac{1}{a^2}-\frac{1}{{\alpha }^2}}{b^2{\beta }^2}=-\frac{(\frac{1}{b^2}-\frac{1}{{\beta }^2})}{a^2{\alpha }^2}\Rightarrow {\alpha }^2+{\beta }^2=a^2+b^2$
note: remember it as formula