Question:

Find the equation(s) of the tangents to $\displaystyle x^2 + 4y^2 = 16$ which pass through

a) the point (2,2)

b) the point (4,6).

Attempted Solution:$\displaystyle y = \pm\sqrt{\frac{-x^2 + 16}{4}}$

$\displaystyle y' = \frac{-x}{4y}$

a) For $\displaystyle P(2,2)$:

Let $\displaystyle P_{tan}$ be parametrized as $\displaystyle (a, f(a)$.

$\displaystyle \frac{2 - \pm\sqrt{\frac{-a^2 + 16}{4}}}{2 - a} = \frac{rise}{run}= m$

What do I do from here? The points given are not points of tangency; they are points that the tangents pass through. So, do I have to set something equal and solve for $\displaystyle a$?