Show by implicit differentiation that the tangent to the ellipse:

x^(2) / a^(2) + y^(2) / b^(2) = 1

*** $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
at the point (x0, y0) is:

[ (x0x) / a^(2) ] + [ y0y / b^(2) ]

*** $\displaystyle \frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1$ . Please do note the correction "=1" done to the above
I got my gradient to be y' = -b^(2)x0 / a^(2)y0 but my tangent line was:

y = [ -b^(2)x0 / a^(2)y0 ] + [b^(2)x0^(2) / a^(2)y0^(2)]

*** $\displaystyle y=\frac{-b^2x_0}{a^2y_0}+\frac{b^2x_0^2}{a^2y_0^2}$ We agree: $\displaystyle y'=\frac{dy}{dx}=-\frac{b^2x}{a^2y}$ and at the point $\displaystyle (x_0, y_0)$ we get what you wrote, but what you say is the tangent line's eq. cannot be correct since, for example, that has no $\displaystyle x-$term in it...! The eq. of the tangent line to the ellipse at this point is in fact: $\displaystyle y-y_0=-\frac{b^2x_0}{a^2y_0}(x-x_0)\Longrightarrow a^2y_0y=-b^2x_0(x-x_0)+a^2y_0$ $\displaystyle \Longrightarrow a^2y_0y+b^2x_0x=b^2x_0^2+a^2y_0^2$ , and now just divide both sides by $\displaystyle a^2b^2$ __and remember__ that $\displaystyle (x_0,y_0)$ is a point on the ellipse...and you'll get what you were asked! Tonio
Im not sure what the problem may be as I simply subbed in the gradient and (x0,y0) points into the y = mx + c equation