# Thread: tangent line to the ellipse-zero in the denominator

1. ## tangent line to the ellipse-zero in the denominator

I need help with this one.

Find the tangent lines to the ellipse x^2 + 2y^2 +4x = 5 at the point (1,0)
I differentiated the equation implicitly to get this: (-2x-4)/4y
And im stuck here. cuz i get a ZERO in the denoinator.
the teacher said something like this is gonna appear on the test, so please help!!!

2. Originally Posted by hsclassofohnine
I need help with this one.

Find the tangent lines to the ellipse x^2 + 2y^2 +4x = 5 at the point (1,0)
I differentiated the equation implicitly to get this: (-2x-4)/4y
And im stuck here. cuz i get a ZERO in the denoinator.
the teacher said something like this is gonna appear on the test, so please help!!!
If you drew the ellipse you'd realise that the given point is a vertex. The equation of the tangent is x = 1.

$x^2 + 2y^2 + 4x = 5 \Rightarrow \frac{(x + 2)^2}{9} + \frac{2y^2}{9} = 1$.