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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.

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    Just spotted something the secound I published it, I have not conciderd the x in the linear equation $y = kx + m$.. Will see if I can solve it. – 2017-02-25
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    Should the derivative actually be a partial derivative, technically? – 2017-02-25
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    I do not know, probably not but it was the only thing I could come up with. lab bhattacharjee came up with a much better solution, but I do not fully understand it. – 2017-02-25

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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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    Sorry I do not really understand, how can you conclude the $xx_1$ and $yy_1$ part from $y'$? – 2017-02-25
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    @Salviati, what is $y'$ at $x_1,y_1$? – 2017-02-25
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    @labbhattacharjee I am stumped too. How does $ y'=-\dfrac{b^2x_1}{a^2y_1}$ imply $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$ – 2018-03-30