Question

Two curves $a_i{x}^2+b_i{y}^2=1;i=1,2$ where $a_1\ne a_2,b_1\ne b_2,a_1,a_2,b_1,b_2\ne 0$ may intersect orthogonally if _____

• A. $a_1{a}_2=b_1{b}_2$
• B. $a_1^{-1}-a_2^{-1}=b_1^{-1}-b_2^{-1}$
• C. $a_1^{-1}+a_2^{-1}=b_1^{-1}+b_2^{-1}$
• D. $a_1{b}_2=a_2{b}_1$

Answer 1

The correct option is B $a_1^{-1}-a_2^{-1}=b_1^{-1}-b_2^{-1}$

The curves are
$a_1{x}^2+b_1{y}^2=1\cdots \left(1\right)$
and $a_2{x}^2+b_2{y}^2=1\cdots \left(2\right)$

Differentiating both equations
${\left(\frac{dy}{dx}\right)}_1=-\frac{a_{1}x}{b_{1}y}$
${\left(\frac{dy}{dx}\right)}_2=-\frac{a_{2}x}{b_{2}y}$

The curves intersect orthogonally.
$\Rightarrow (-\frac{a_{1}x}{b_{1}y})(-\frac{a_{2}x}{b_{2}y})=-1$

$\Rightarrow a_1{a}_2{x}^2=-b_1{b}_2{y}^2\cdots \left(3\right)$

$\left(1\right)-\left(2\right)$ gives
$(a_1-a_2)x^2+(b_1-b_2)y^2=0$
$\Rightarrow (a_1-a_2)\frac{-b_1{b}_2{y}^2}{a_1{a}_2}+(b_1-b_2)y^2=0(\text{From (3)})$
$\Rightarrow \frac{a_1-a_2}{a_1{a}_2}=\frac{b_1-b_2}{b_1{b}_2}$
$\Rightarrow \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{b_2}-\frac{1}{b_1}$
$\Rightarrow \frac{1}{a_1}-\frac{1}{a_2}=\frac{1}{b_1}-\frac{1}{b_2}$