Thank you for reporting, we will resolve it shortly
Q.
The normal to a curve at $P(x, y)$ meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a
Differential Equations
Solution:
Slope of tangent $=\frac{dy}{dx}$
$\therefore $ Slope of normal $=-\frac{dx}{dy}$
Thus, the equation of normal is
$Y-y=-\frac{dx}{dy}(X-x)$
This meets x-axis (y = 0), where
$-y=-\frac{dx}{dy}(X-x)$ or $X=x+y \frac{dy}{dx}$
$\therefore \,G$ is $(x+y \frac{dy}{dx}, 0)$
$\therefore OG=2x$
$\therefore x+y\frac{dy}{dx}=2x$
or $y \frac{dy}{dx}=x$ or $y\,dy=xdx$
Integrating, we get $\frac{y^{2}}{2}=\frac{x^{2}}{2}+\frac{C}{2}$
or $y^{2}-x^{2}=c$, which is a hyperbola