Question

If the curves $x^2+p{y}^2=1$ and $q{x}^2+y^2=1$ are orthogonal to each other, then

• A. $p-q=0$
• B. $\frac{1}{p}-\frac{1}{q}=2$
• C. $\frac{1}{p}+\frac{1}{q}=-2$
• D. $\frac{1}{p}+\frac{1}{q}=2$

Answer 1

Answer: (D)

$\frac{1}{p}+\frac{1}{q}=2$

$x^2+p{y}^2=1$ ...(2)
$q{x}^2+y^2=1$ ...(3)
subtracting,
$x^2(1-q)+y^2(p-1)=0$
$\Rightarrow \frac{x^2}{y^2}=\left(\frac{p-1}{q-1}\right)$ ...(1)
Differentiating eqn (2),
$2x+2py\frac{dy}{dx}=0$
$\frac{dy}{dx}{|}_2=\frac{-x}{py}$
Differentiating eqn (3)
$2qx+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}{|}_3=\frac{-qx}{y}$
Condition for orthogonality,
$\frac{-x}{py},\frac{-qx}{y}=-1\Rightarrow \frac{q{x}^2}{p{y}^2}=-1$ ...(4)
from eq^n (1) and (4)
$\frac{q}{p}\left(\frac{p-1}{q-1}\right)=-1$
$\Rightarrow qp-q=-[pq-p]$
$\to qp-q=-pq+p$
$\Rightarrow 2qp=p+q$
$\Rightarrow 2=\frac{1}{p}+\frac{1}{q}$