Answer to Question #157497 in Differential Geometry | Topology for Bawe

Question

Prove that the equation of the tangent and normal to the ellipse (x^2/a^2) + (y^2/b^2) = 1 at the point P (acosu , bsinu) are respectively
bxcosu + aysinu = ab and
axsinu + bycosu = (a^2 - b^2)sinucosu.

Answer 1

The equation of tangent and normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ at the point } (x_1,y_1) \text{ are } \frac{x_1x}{a}+\frac{y_1y}{b}=1 \text{ and } a^2y_1x-b^2x_1y-(a^2-b^2)x_1y_1=0 respectively.\\ \text{At } P(a\cos u, b\sin u), \text{ the equation of tangent will be;}\\ \frac{ax\cos u}{a}+\frac{by\sin u}{b}=1\\ x\cos u+y\sin u=1\\ \text{Also, the equation of normal will be; }\\ a^2(b\sin u) x-b^2(a\cos u) y=(a^2-b^2)(ab\sin u \cos u) \\ a^2bx\sin u-ab^2y\cos u=(a^2-b^2)(ab\sin u \cos u) \\ \text{Divide through by ab}\\ ax\sin u-by\cos u=(a^2-b^2)\sin u\cos u$