Question

The equation of the normal at the point (2,3) on the ellipse 9x ^2+ 16y ^2=180 is

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The equation of the normal is 3y - 8x + 7 = 0.

Given: The equation of the ellipse, $9x^{2} +16y^{2} =180$ and

             The point ( 2,3)

To find:

         The equation of the normal.

Solution:

        We know,

            $\frac{dy}{dx }$ = Slope of tangent

and slope of tangent = negative inverse of the slope of normal

                                                                          ║∵ the normal ⊥ tangent ║

Differentiating with respect to x,

           We get, $18x+32y\frac{dy}{dx} =0$

At the point (2,3)

         The equation will be, $36+96\frac{dy}{dx} =0$

                          ⇒ $\frac{dy}{dx} = -36/96$

   ∴ Slope of tangent = -3/8

   ∴ The slope of normal = 8/3

⇒ Equation of the normal will be,

                               ( y - 3 ) = 8/3 ( x - 3 )

                    ⇒  3y - 8x + 7 = 0 will be the equation of the normal.