Question

The equation of a curve passing through $(1,0)$ for which the product of the abscissa of a point $P$ and the intercept made by a normal at $P$ on the $x$-axis equals twice the square of the distance of the point $P$ from the origin $O$, is

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

Answer 1

The correct option is A $x^2+y^2=x^4$

Tangent at point P is $Y-y=-\frac{1}{m}(X-x)$where $m=\frac{dy}{dx}$.
Let $Y=0$. Then $X=my+x$

According to question, $x(my+x)=2(x^2+y^2)$
or $\frac{dy}{dx}=\frac{x^2+2y^2}{xy}$ (homogeneous)
Putting $y=vx$, we get
$v+x\frac{dv}{dx}=\frac{1+2v^2}{v}$
or $x\frac{dv}{dx}=\frac{1+2v^2}{v}-v=\frac{1+v^2}{v}$
or $\int \frac{vdv}{1+v^2}=\int \frac{dx}{x}$
or $\frac{1}{2}\log (1+v^2)=\log x+\log c,c>0$
or $x^2+y^2=c{x}^4$
Also, it passes through (1, 0). Then $c=1$.