Question

Find the equation of tangent and normal to the ellipse $9x^{2} +16y^{2} =144$ at the end of the latus rectum in the first quadrant.

Answer 1

The normals at the ends of latus rectum (in first quadrant) of the ellipse 9x2 + 16y2 = 144

Equation of the ellipse 9x2 + 16y2 = 144can be written as

x216+y29=1
a=4, b=3

e2  = 1 −b2a2 = 1−916 = 716

In the first quadrant, the co-ordinates at the end of latus rectum will be (ae,b1−e2‾‾‾‾‾‾√)equaling (7,‾‾√94/)
9x2 + 16y2 = 144

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

or dy/dx=-9x/16y

Slope of tangent at (7,‾‾√94/) is -7‾√/4
Slope of normal becomes 47√/
So, the equation of normal will be

(y−94)=(47√)(x−7‾√)

Simplifying we get

16x-47‾√y=77‾√
There will be only one normal in first quadrant at the end of first quadrant