Show that the equation of the normal at the point $x = a\cos(t)$, $y = b\sin(t)$ of the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ is $$\frac{2a^2 - b^2}{a}$$
Hi, I am not sure how to proceed on this question, i have got as far as:
$$x\ a\ \sin(t) - y\ b\ \cos(t) = (a^2 - b^2)\sin(t)\cos(t)$$
By using $y - y1 = m(x - x1)$ and the gradient from the derivative of the ellipse equation. Please can you help
You know that the normal of the region $F(x,y) = C$ at the point $(x_0, y_0)$ has direction $\nabla F(x_0, y_0)$.
Here $F(x,y) = x^2/a^2 + y^2/b^2$, hence the direction is $ (x_0/a^2, y_0/b^2) $ and the complete parametric equation is $$ M(t) = (x_0 + x_0/a^2 t, y_0 + y_0/b^2 t) $$ or, in implicit form (just get rid of $t$): $$ y/y_0 - 1 = \frac{a^2}{b^2} (x/x_0 - 1) $$