Question: How would I go about solving the following question?

> 
> Find the equations of the lines that are tangent to the ellipse $x^2 + 4y^2 = 16$ and that also pass through the point $(4,6)$
> 
> 
> 

Please provide some hints/clues

Thanks!

Answer: 
Equation of any line passing through $(4,6)$ can be written as $$\frac{y-6}{x-4}=m$$ where $m$ is the gradient.

So, $y=mx+6-4m$

Let us find the intersection of the line with the ellipse.

So, $$x^2+4(mx+6-4m)^2=16$$
$$\implies (1+4m^2)x^2+8xm(6-4m)+4(6-4m)^2-16=0$$

which is a quadratic equation in $x$

For tangency, both the root should be same, i.e., the discriminant must be $0$ which will give us the value of $m$

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Alternatively, from the Article $(278)$ of [this](http://archive.org/details/elementsofcoordi00loneuoft), the equation of the pair of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ from $P(x\_1,y\_1)$ will be $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)\left(\frac{x\_1^2}{a^2}+\frac{y\_1^2}{b^2}-1\right)=\left(\frac{xx\_1}{a^2}+\frac{yy\_1}{b^2}-1\right)^2$$
Equation of tangent to ellipse $$x^2/a^2+y^2/b^2=1$$ at point $T(x\_0,y\_0)$ is 
$$xx\_0/a^2+yy\_0/b^2=1$$