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

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> 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: 
1) Find slopes of the tangents to the ellipse at any point on the ellipse. Implicit differentiation helps here, i.e.

$$2 x + 8 y \frac{dy}{dx} = 0$$

2) Find the point(s) on the ellipse whose tangent passes through $(4,6)$. The equation of a line is

$$y-y\_0 = m (x - x\_0)$$

where $m$ is the slope at the point $(x,y)$ on the ellipse and $(x\_0,y\_0)$ is the point $(4,6)$. You end up with an equation like

$$y-6 = \frac{dy}{dx} (x-4)$$

Plug in the derivative for the slope; you will get 2 equations and 2 unknowns (the other equation being the equation for the ellipse). Solution will get you the point(s), and thus the slope(s) and the equation(s) of the lines.

So you can check your results, I get two points: $(4,0)$ (obvious in hindsight) and $(-16/5,6/5)$.
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 $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)\left(\frac{x\_1^2}{a^2}+\frac{y\_1^2}{b^2}-1\right)=\left(\frac{xx\_1}{a^2}+\frac{yy\_1}{b^2}-1\right)^2$$