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?
On
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.
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