Show the condition that the curves `a x^2+b y^2=1` and `Ax^2+By^2=1` should intersect orthogonally is `1/a-1/b=1/A-1/Bdot`

Options

A. `(1)/(a)+(1)/(A)=(1)/(b) + (1)/(B)`
B. `(1)/(a)-(1)/(A)=(1)/(b) - (1)/(B)`
C. `(1)/(a)+(1)/(b)=(1)/(B) - (1)/(A)`
D. `(1)/(a)+(1)/(b)=(1)/(A) + (1)/(B)`

Answer

B

Explanation

In illustration 4, we have seen that the curves `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(c^(2))+(y^(2))/(d^(2))=1` intersect orthogonally , if `a^(2) -b^(2)=c^(2)-d^(2).`
So, given curve will intersect orthogonally, if `(1)/(a)-(1)/(b)=(1)/(A) - (1)/(B)`.