<s>[INST] Ellipse cutting orthogonally 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"?
 [/INST] 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, $m_1m_2=-1$
</s>