Question

Find the co-ordinates of the points on the ellipse $x^2+2y^2=9$ at which tangent has slope $\frac{1}{4}$. Also find the equation of normal.

Answer 1

$x^2+2y^2=9$
$\therefore 2x+4y\frac{dy}{dx}=0$
$\therefore x+2y\frac{dy}{dx}=0$
$\therefore \frac{dy}{dx}=-\frac{x}{2y}$
Now slope of tangent $=\frac{1}{4}$
$\therefore -\frac{x}{2y}=\frac{1}{4}$
$\therefore x=-\frac{y}{2}$ $\left(1\right)$
Put $x=-\frac{y}{2}$ in equation $x^2+2y^2=9$
$\therefore \frac{y^2}{4}+2y^2=9$
$\therefore y^2+8y^2=36$
$\therefore 9y^2=36$
$\therefore y^2=4$
$\therefore y=\pm 2$
Substitute this value in result $\left(1\right)$
$\therefore x=\frac{\pm 2}{2}$
$\therefore x=\pm 1$
$\therefore$ Required points means co-ordinates of the point of contact $(1,2)$ and $(-1,2)$
Now slope of tangent $\frac{dy}{dx}=\frac{x}{2y}$
$\therefore {\left(\frac{dy}{dx}\right)}_{P(1,2)}=\frac{1}{4}$
$\therefore$ Slope of Normal $=-4$
$\therefore$ We get equation of Normal using equation
$y=y_1=m(x-x_1)$
$\therefore y+2=-4(x-1)$
$\therefore y+2=-4x+4$
$\therefore 4x+y=2$ .$\left(1\right)$
$\therefore$ Now equation of normal at point
$(-1,2)y-2=-4(x+1)$
$\therefore y-2=-4x-4$
$\therefore 4x+y=-2$ .$\left(2\right)$
$\therefore$ Result $\left(1\right)$ and $\left(2\right)$ shows equation of Normal.