Question

The normal at any point $P(x,y)$ of a curve meets the x-axis at $Q$ and $N$ is the foot of the ordinate at $P$.
If $NQ=\frac{x(1+y^2)}{1+x^2},$ then equation of such curve, given that it passes through the point $(3,1)$ is:

• A. $x^2-y^2=8$
• B. $x^2+2y^2=11$
• C. $x^2-5y^2=4$
• D. $x^2+3y^2=7$

Answer 1

The correct option is C $x^2-5y^2=4$

The equation of the normal at P, can be written as $Y-y=\frac{-1}{\left(dy/dx\right)}(X-x)$
The normal to the curve meets the x-axis at $Q(x+y\frac{dy}{dx},0)$.
$N$ is the foot of the ordinate. Hence, $N$ has the coordinates $(x,0)$.
$NQ=y\frac{dy}{dx}=\frac{x(1+y^2)}{1+x^2}$
$\int \frac{y}{1+y^2}dy=\int \frac{x(1+y^2)}{1+x^2}dx$
$\frac{1}{2}\ell n(1+y^2)=\frac{1}{2}\ell n(1+x^2)+\frac{1}{2}\ell nc$
$(1+y^2)=c(1+x^2)$
The given curve passes through $(3,1)$.
Hence, $2=c\left(10\right)$
$\Rightarrow c=\frac{1}{5}$
$\therefore 5(1+y^2)=1+x^2$
$x^2-5y^2=4$