Answer to Question #220625 in Calculus for Stacey

Question

The equation of the ellipse is given as $(x/a)^2+(y/b)^2 =1$

find the equation of the tangent at (x₀, y₀)

Answer 1

$\frac{d}{dx}((\frac{x}{a})^2+(\frac{y}{b})^2)=\frac{d}{dx}(1).$

$2\frac{x}{a^2}+2\frac{y}{b^2}y'=0.$

$y'=-\frac{b^2}{a^2}\frac{x}{y}.$

$y'(x_0,y_0)=-\frac{b^2}{a^2}\frac{x_0}{y_0}.$

Tangent line: $y-y_0=-\frac{b^2}{a^2}\frac{x_0}{y_0}(x-x_0).$

Or $y=-\frac{b^2}{a^2}\frac{x_0}{y_0}x+\frac{b^2}{a^2}\frac{x_0}{y_0}x_0+y_0.$