Ellipse cutting orthogonally

Question

If the curves $ax^2+by^2=1$ and $a'x^2+b'y^2=1$ cut orthogonally, then :

A)$\displaystyle \frac{1}{b}+\frac{1}{b'}=\frac{1}{a}+\frac{1}{a'}$

B)$\displaystyle \frac{1}{b}-\frac{1}{b'}=\frac{1}{a}-\frac{1}{a'}$

C)$\displaystyle \frac{1}{b'}-\frac{1}{b}=\frac{1}{a}-\frac{1}{a'}$

D)$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{a'}+\frac{1}{b'}=0$

I tried solving it but failed. My doubt is "Can two ellipses having common centre ever cut orthogonally"?

Answer 1

HINT:

Let the two curves intersect at $(h,k)$

Solve for $(h,k)$ in terms $a,b,a',b'$

Now, the gradient $(m_1)$ of $ax^2+by^2=1$ is $-\dfrac{ah}{bk}$

As the two curves cut orthogonally, $m_1m_2=-1$

Answer 2

To answer OP's specific doubt:

Two axis-aligned ellipses with common centre cannot intersect orthogonally. However, the point is that the coefficients are not necessarily positive, so one curve (or both) could be a hyperbola rather than an ellipse.