Write an equation for the tangent line to the ellipse x^2 + 4y^2 = 8 at the point (-2,1)?

Answer 1

I assume you are in calculus. If you are in precalculus, ignore this answer; you have to do it the hard way!

To get the tangent line, you just need a point and the slope. You have the point. Get the slope. The slope at any point on this curve is dy/dx. Implicitly differentiate the equation wrt x:

2x + 8y * dy/dx = 0
dy/dx = -2x / 8y

At (-2, 1) dy/dx = 4/8 = 1/2, so the equation of the line is

y - 1 = 1/2 (x - -2)
y = 1/2*x +2

Answer 2

x^2 + 4y^2 = 8
Using implicit differentiation we get
2x + 8y dy/dx = 0
So at (-2, 1) we get
-4 + 8 dy/dx = 0
and hence dy/dx = 1/2.
So the equation is
(y - 1) = 1/2 (x + 2)
<=> y = x/2 + 2.