Question #220625

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

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


1
Expert's answer
2021-07-26T16:31:00-0400

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

2xa2+2yb2y=0.2\frac{x}{a^2}+2\frac{y}{b^2}y'=0.

y=b2a2xy.y'=-\frac{b^2}{a^2}\frac{x}{y}.

y(x0,y0)=b2a2x0y0.y'(x_0,y_0)=-\frac{b^2}{a^2}\frac{x_0}{y_0}.

Tangent line: yy0=b2a2x0y0(xx0).y-y_0=-\frac{b^2}{a^2}\frac{x_0}{y_0}(x-x_0).

Or y=b2a2x0y0x+b2a2x0y0x0+y0.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.


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