Question

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

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

Answer 1

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

For ellipse slope of tangent $=\frac{-x{b}^2}{y{a}^2}$
For hyperbola $t_2=\frac{x{m}^2}{y{l}^2}$
As they intersect
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{x^2}{l^2}-\frac{y^2}{m^2}$
$y^2(\frac{1}{b^2}+\frac{1}{m^2})=x^2(\frac{1}{l^2}-\frac{1}{a^2})$
They intersect arthogonally $\frac{x^2}{y^2}\frac{b^2{m}^2}{a^2{l}^2}=1$
$\therefore t_{1}t2=-1$
$\frac{l^2{a}^2}{b^2{m}^2}\frac{(m^2+b^2)}{(a^2-l^2)}\frac{b^2{m}^2}{a^2{l}^2}=1$
$\therefore m^2+b^2=a^2-l^2$
$\therefore a^2-b^2=l^2+m^2$