Question

find the normals at the ends of latus rectum (in first quadrant) of the ellipse 9x^2 +16y^2 =144

Answer 1

Dear Student,

The normals at the ends of latus rectum (in first quadrant) of the ellipse $9x^2+16y^2=144$

Equation of the ellipse $9x^2+16y^2=144$ can be written as

$\frac{x^2}{16}+\frac{y^2}{9}=1$
a=4, b=3

$e^2=1-\frac{b^2}{a^2}=1-\frac{9}{16}=\frac{7}{16}$

In the first quadrant, the co-ordinates at the end of latus rectum will be $\left(ae,b\sqrt{1-e^2}\right)$ equaling $\left(\sqrt{7,}\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)$
$9x^2+16y^2=144$

Differentiating we get 18x+32y*dy/dx=0

or dy/dx=-9x/16y

Slope of tangent at $\left(\sqrt{7,}\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)$ is -$\sqrt{7}/4$
Slope of normal becomes $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$\sqrt{7}$}\right.$
So, the equation of normal will be

$\left(y-\frac{9}{4}\right)=\left(\frac{4}{\sqrt{7}}\right)\left(x-\sqrt{7}\right)$

Simplifying we get

16x-4$\sqrt{7}$y=7$\sqrt{7}$
There will be only one normal in first quadrant at the end of first quadrant

Regards