Question

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ and $x^2-y^2=c^2$ cut at right angles, then :

• A. $a^2+b^2=2c^2$
• B. $a^2-b^2=2c^2$
• C. $a^2{b}^2=2c^2$
• D. None of these

Answer 1

The correct option is B $a^2-b^2=2c^2$

We can easily determine the coordinates of the intersection points of the curves.
The coordinates are,
$(x,y)=(\sqrt{\frac{a^2(b^2+c^2)}{a^2+b^2}},\sqrt{\frac{b^2(a^2-c^2)}{a^2+b^2}})$
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,
$\frac{2x}{a^2}+\frac{2y{y}^{\prime }}{b^2}=0$
$y^{\prime }=-\frac{b^{2}x}{a^{2}y}$ ...(i)
$2x-2y{y}^{\prime }=0$
$\frac{x}{y}=y^{\prime }$ ...(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
$\frac{b^2}{a^2}=\frac{b^2(a^2-c^2)}{a^2(b^2+c^2)}$
$b^2+c^2=a^2-c^2$
$a^2-b^2=2c^2$