The coordinates are,
$(x, y)=\left(\sqrt{\frac{a^2\left(b^2+c^2\right)}{a^2+b^2}}, \sqrt{\frac{b^2\left(a^2-c^2\right)}{a^2+b^2}}\right)$
Since the curves intersect at right angles, then their slopes at that point will be perpendicular.
That is, the product of slopes will be $-1$.
Differentiating both the equations,
$\frac{2 x}{a^2}+\frac{2 y y^{\prime}}{b^2}=0$
$y^{\prime}=-\frac{b^2 x}{a^2 y} \ldots \text { (i) }$
$2 x-2 y^{\prime}=0$
$\frac{x}{y}=y^{\prime} \ldots \text { (ii) }$
Their product is $-1$.
Hence, $-\frac{x}{y} \frac{b^2 x}{a^2 y}=-1$
$\therefore \frac{b^2 x^2}{a^2 y^2}=1$
Substituting the point we get
$\frac{b^2}{a^2}=\frac{b^2\left(a^2-c^2\right)}{a^2\left(b^2+c^2\right)}$
$b^2+c^2=a^2-c^2$
$a^2-b^2=2 c^2$
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