Question

The normal at a point P(x, y) of a curve meets the x-axis at Q and N is the foot of the perpendicular from P on x-axis, If $NQ=\frac{x(1+y^2)}{1+x^2}$, then equation of the curve passing through (3, 1) is

• A. $x^2-y^2=8$
• B. $x^2+y^2=4$
• C. $x^2+y^2=10$
• D. $x^2-5y^2=4$

Answer 1

The correct option is D

$x^2-5y^2=4$

Equation of normal at P(x, y) is
$Y-y=-\frac{dx}{dy}(X-x)$
It meets the x-axis when $X=y\frac{dy}{dx}+x$
$\Rightarrow OQ=x+y\frac{dy}{dx}$
Also $ON=x\Rightarrow NQ=y\frac{dy}{dx}$
$\Rightarrow x\frac{(1+y^2)}{1+x^2}=y\frac{dy}{dx}\Rightarrow \frac{xdx}{1+x^2}=\frac{ydy}{1+y^2}$
$\Rightarrow \int \frac{2xdx}{1+x^2}=\int \frac{2ydy}{1+y^2}$
or $log(1+x^2)=log(1+y^2)+C\Rightarrow \frac{1+x^2}{1+y^2}=k$
As the curves passes through (3, 1) therefore k = 5
Hence the equation is $x^2-5y^2=4$