Tangents of ellipse going through a point outside of it

Question

I have an ellipse: 5x^(2) + 9y^(2) = 45; and a point M = [0; -3]; how do I find tangents of said ellipse that go through the point M? Please, help.

Answer 1

Suppose the tangent line going through $M$ touches the ellipse at $(x_0, y_0)$. Then the equation of tangent line on that point is $$5x_0x+9y_0y = 45$$ Since this line goes through $M=(0, 3)$, We have $y_0=-\frac{5}{3}$.

Also because $(x_0, y_0)$ is on the ellipse, it satisfies $5x^2+9y^2=45$, so plugging in $y_0 = \frac{5}{3}$ would give $x_0 = \pm 2$.

Thus the final answer is $$\pm 2x-3y=9$$

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A bit of additional explanation for how I got the tangent line: Calculating the derivative (implicit function) gives $$10x+18y\frac{dy}{dx} = 0$$ thus the slope of the tangent line on $(x_0, y_0)$ would be $$\frac{dy}{dx} = -\frac{5x_0}{9y_0}$$ So the equation would be $$y-y_0 = -\frac{5x_0}{9y_0}(x-x_0)$$ and simplifying with $5x_0^2+9y_0^2=45$ will give $$5x_0x + 9y_0y=45$$

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This is actually a useful technique for calculating tangent lines on conic sections. Generally, tangent line to $Ax^2+By^2=C$ on $(x_0, y_0)$ is $$Ax_0x+By_0y=C$$ (Just replace $x^2$ with $x_0x$, $y^2$ with $y_0y$)

Answer 2

We have given $$M(0;-3)$$ and our tangentline has the form $$y=mx-3$$ plugging this in the equation of our Ellipse we get $$5x^2+9(mx-3)^2=45$$, now you must compute $m$ such that this equation has only one solution. This is $$45m^2-20=0$$