Question

A point on the ellipse x^2+3y^2=37 at which the normal is parallel to the line 6x-5y=2

Answer 1

Answer: The points are (5,2) and (-5,-2) on the ellipse $x^2+3y^2=37$ at which the normal is parallel to the line 6x-5y=2.

Explanation:

The equation of ellipse is

$x^2+3y^2=37$

Differentiate w.r.t x.

$2x+6y(\frac{dy}{dx}) =0$

$\frac{dy}{dx}=\frac{-x}{3y}$ =0[/tex]

The slope of normal is $-\frac{dx}{dy}$.

$-\frac{dx}{dy}=-(\frac{-3y}{x})=\frac{3y}{x}$

Slope of normal is $\frac{3y}{x}$.

The normal is parallel to the line $6x-5y=2$, therefore he slope of this line and the slope of normal is equal.

$6x-5y=2$

$6-5\frac{dy}{dx} =0$

$\frac{dy}{dx}=\frac{6}{5}$

Slope of lines is $\frac{6}{5}$.

$\frac{3y}{x}=\frac{6}{5}$

$15y=6x$$5y=2x$

$x=\frac{5y}{2}$  .....(1)

Put this value in the equation of ellipse.

$(\frac{5y}{2})^2+3y^2=37$

$\frac{25y^2}{4}+3y^2=37\\\frac{25y^2+12y^2}{4}=37$

$\frac{37y^2}{4}=37$

$y^2=4$

$y=\sqrt{4}$

So, the value of y is either -2 or 2.

Put x=2 in equation (1) we get y=5.

Put x=-2 in equation (2) we et y=-5.

Therefore, the points are (5,2) and (-5,-2) on the ellipse $x^2+3y^2=37$ at which the normal is parallel to the line 6x-5y=2.