Please how do I go by this?

Question

The perpendicular OY is drawn from the centre O of the ellipse x^2/a^2 + y^2/b^2 = 1 to

any tangent. Prove that the locus of Y is (x^2 + y^2)^2 = a^2x^2 + b^2y^2

Answer 1

First, we need to find the slope of the tangent, which is dy/dx.

To find dy/dx for the ellipse, we need to use implicit differentiation. So, we get,

2x/a²+(2y/b²)(dy/dx)=0

dy/dx=-xb²/(ya²)

The slope of the perpendicular to the tangent is the negative reciprocal of dy/dx

So that is ya²/(xb²)

The line OY is drawn from the center O at the origin.

Now, lets assume the tangent to the ellipse is at a point P given by (p,q). Let the coordinate of Y be (x,y).

The slope of the tangent is pb²/(qa²) and the equation of the tangent is y-q=(pb²/(qa²))(x-p)

The slope of the line OY is qa²/pb² and the equation of OY is y=(qa²/pb²)x

The point of intersection of the two lines is the point Y.

Once you get the coordinates of the point of intersection, plug the x and y values into the equation of OY.

You will get an equation connecting p and q. That is the locus of the point Y, except you need to change (p,q) to (x,y) to get the answer.