Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$
and pass through $(4,6)$.
I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find the other tangents. Any help is appreciated.
Best Answer
You need a line that passes through the point $(4,6)$ and that touches the ellipse at just one point. The vertical line does that and you've already found it. Obviously there is exactly one other tangent line (and if that's not obvious to you, then draw the picture and look at it!).
Nonvertical lines passing through the point $(4,6)$ have equation $y-6=m(x-4)$.
That implies $y=mx-4m+6$, so we can put $mx-4m+6$ in place of $y$:
$$ \frac{x^2}{16} + \frac{(mx-4m+6)^2}{4} = 1. $$ This is equivalent to $$ \underbrace{(1+4m)}x^2 + \underbrace{-8m(4m-6)}\ x + \underbrace{4(4m-6)^2 -16} = 0. $$
This equation is quadratic in $x$. We therefore want a quadratic equation with exactly one solution. A quadratic equation $ax^2+bx+c=0$ has exactly one solution precisely if its discriminant $b^2-4ac$ is $0$. So we have $$ b^2-4ac = \underbrace{64m^2(4m-6)^2 - 4(1+4m)(4(4m-6)^2-16) = 0}. $$ Now we only need to solve this last equation for $m$.