<s>[INST] Show that the tangent intercept the the ellipse in thease positions I have been stuck on this question for a couple of hours and can not figure it out, it states:

Consider the tangent line to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}= 1$ at the point $(p,q)$ in the first quadrant. Show that the tangent line has $x$-intercept $\dfrac{a^2}{p}$ and $y$-intercept $\dfrac{b^2}{q}$.

How I tried to solve this problem was to start differentiating the function with respect of $x$.
$$\frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right) = 0 \implies \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y'= -\frac{xb^2}{ya^2}$$
Now I want to know what I can substitute $y$ for, so I solve it in the first equation.
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}= 1 \implies \ldots \implies y = \frac{\sqrt{a^2b^2-x^2b^2}}{a}$$
I substitute $y$ to this in the second equation.
$$y'= -\frac{2xb^2}{2a^2 \left( \dfrac{\sqrt{a^2b^2-x^2b^2}}{a} \right)} \implies \ldots \implies y' = -\frac{xb\sqrt{a^2-x^2}}{a(a^2-x^2)}$$
So now I know that the tangent line have the equation $y=-\dfrac{xb\sqrt{a^2-x^2}}{a(a^2-x^2)} + m$ where $m$ is the intersection in the $y$-axis. I know that I have a point $(p,q)$ witch are constants, so I solve $m$.
$$m = q + \frac{pb\sqrt{a^2-p^2}}{a(a^2-p^2)}$$
Now I have the equation for the tangent line I was after.
$$y = -\frac{xb\sqrt{a^2-x^2}}{a(a^2-x^2)} + q + \frac{pb\sqrt{a^2-p^2}}{a(a^2-p^2)}$$
But, here is the problem. If I solve $y(p)$, I get that $y(p) = q$, which make sense. But that was not really what I was after.
Have I misinterpreted the question or can any one of you spot my error? Personally I do not know what else I can do except what I have already done.
 [/INST] As $y'=-\dfrac{b^2x}{a^2y}$
So, the equation of the tangent at $x_1,y_1$
$$\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1\iff\dfrac x{\dfrac{a^2}{x_1}}+\dfrac b{\dfrac{b^2}{y_1}}=1$$
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