I've been stuck on this problem for a while now. Not quite sure how to get at it. I've tried finding the derivative of the equation and using point slope form but cant get it to look like the defined tangent line. Also, a and b are both constants.
Show that $(x_0,y_0)$ is the point at which
$$\frac{x_0}{a^2}x + \frac{y_0}{b^2}y = 1$$
is tangent to the curve
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$
On
We need to show that (1) $(x_0,y_0)$ lies on the lines and that (2) the slope of the line is $f'(x_0)$.
The first condition is obvious, since substituting $(x_0,y_0)$ in for $(x,y)$ yields $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ which is true because $(x_0,y_0)$ lies on the ellipse.
To find the slope, we differentiate implicitly with respect to $x$. This yields
$$ \frac{2x}{a^2}+(\frac{2y}{b^2})(\frac{dy}{dx})=0 $$ So $\frac{dy}{dx}=-\frac{xb^2}{ya^2}$. But the slope of the given line is $-\frac{x_0b^2}{y_0a^2}$, which is exactly the slope of the tangent line at the point $(x_0,y_0)$, so this line is the equation of the tangent line at $(x_0,y_0)$.
Well, the basic assumption must be that the given point is on the ellipse, which means
$$\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1\iff x_0^2b^2+y_0^2a^2=a^2b^2$$
Now the slope of tangent line at any point on the ellipse is given by the value of the derivative at the point, so differentiating implicitly the ellipse's equation:
$$\frac{2x}{a^2}dx+\frac{2y}{b^2}dy=0\implies \frac{dy}{dx}=-\frac{b^2}{a^2}\frac xy$$
which is the slope of the given line at the given point...