Question

How do you find the equations of both the tangent lines to the ellipse `x^2 + 4y^2 = 36` that pass through the point (12,3)?

Answer 1

`((y = 3),(y=3+2/3(x-12)))`

Explanation:

The tangent space to the ellipse is obtained calculating the implicit derivative of `f(x,y) = x^2+4y^2-36=0`

The total derivative is calculated as `df = f_xdx+f_ydy=0->dy/(dx)=-f_x/f_y`
So `dy/(dx) = -x/(4y)`. The support point to the tangent lines to the ellipse is `p_0=(p_x^0,p_y^0)=(12,3)` and the tangents to the ellipse are giving by
`r_t->(y-p_y^0)/(x-p_y^0)=-x/(4y)`

Solving
`(( x^2+4y^2-36=0),(y =p_y^0-x/(4y)(x-p_y^0)))`
for `x,y` we obtain
`((x = 0, y = 3), (x = 24/5, y = -9/5))`

Those are the tangency points.
The tangent lines are:
`r_1->y = 3+((3-3)/(0-12))(x-12)`
`r_2->y = 3 +( (-9/5-3)/(24/5-12))(x-12)`