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This question already has an answer here:

First: This is not a question for an assignment!

I have an ellipse given by the equation \begin{equation} \frac{x^2}{2^2} + \frac{y^2}{1^2} = 1 \end{equation} and the intersection point $(0,4)$ of two tangents to this ellipse.

How do I determine the two tangent points to the ellipse for the aforementioned tangents?

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marked as duplicate by Henning Makholm, Juniven, Andres Mejia, Nosrati, hardmath Apr 7 '17 at 17:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ My immediate plan would be to strech the coordinate system such that the ellipse becomes an circle (which does not affect straight lines or tangency), find the tangents there and then transform back to the original coordiantes. $\endgroup$ – Henning Makholm Apr 7 '17 at 9:22
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    $\begingroup$ See and read this $\endgroup$ – Juniven Apr 7 '17 at 9:27
  • $\begingroup$ @ΘΣΦGenSan Thanks. I'll look into that. $\endgroup$ – Svend Tveskæg Apr 7 '17 at 9:28
  • $\begingroup$ @Svend, my suggestion corresponds to what anon's answer to the duplicate describes. $\endgroup$ – Henning Makholm Apr 7 '17 at 9:29
  • $\begingroup$ @HenningMakholm Hej Henning. Har du tid til at tage et kig på mit andet spørgsmål på math.stackexchange.com/questions/2222258/…? $\endgroup$ – Svend Tveskæg Apr 7 '17 at 14:14
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Samjoe gave a nice succinct answer. But I’d like to elaborate a little:

To get the slope of the tangent line, which is a derivative, we can use implicit differentiation: The ellipse is $$\frac{x^2}{4}+y^2=1$$ Now find the derivative: $$\frac{2x}{4}+2y\,y^\prime=0$$ $$2y\,y^\prime=-\frac{x}{2}$$ $$y^\prime=-\frac{x}{4y}$$ Now let's denote our tangent point on the ellipse as $(x_o,\,y_o)$ , and, denoting the slope as $k$, we get: $$k=-\frac{x_o}{4y_o}$$ So our general equation of the tangent line to the ellipse is $$y-y_o=-\frac{x_o}{4y_o}(x-x_o)$$ Now we multiply the both sides of our equation by $y_o$: $$y\,y_o-y_o^2=-\frac{x_o\,x}{4}+\frac{x_o^2}{4}$$ Now we rearrange the terms: $$\frac{x_o\,x}{4}+y\,y_o=\frac{x_o^2}{4}+y_o^2$$ The right hand side of the equation $\frac{x_o^2}{4}+y_o^2=1\;$ because the point $(x_o,y_o)$ lies on the ellipse. So we get $$\frac{x_o\,x}{4}+y\,y_o=1$$ This is the equation of our tangent line/lines. All we need to do now is to plug $x=0$ and $y=4$. We should not confuse the point/points $(x_o,y_o)$ on the ellipse and the point $(0,4)$ which is outside. So we get $$0\cdot x_o+4\,y_o=1$$ and $\quad y_o=1/4.\quad$ Now it's trivial to find $\;x_o.\;$ We just plug $\; y_o=1/4\;$ into the equation of the

ellipse to get the values of $\; x_o$: $\quad x_o=±\sqrt{15}/2$

So, the two tangent points are $\;(-\sqrt{15}/2,\;\,1/4)\;$ and $\quad(\sqrt{15}/2,\;\,1/4)$

Hope it was helpful

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There's a standard result regarding conics, $T=0$, which gives the equation of chord of contant.

$$\frac{xx_1}{2^2} + \frac{yy_1}{1^2} = 1$$

where $(x_1, y_1)$ is the point of intersection of tangents. Solve this equation with ellipse equation to obtain the two points to which tangent was drawn.

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