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

Answer 1

x^2 + 4y^2 = 8
Implicit differentiation gives
2x + 8y dy/dx = 0 => dy/dx = -x/4y
At (-2, 1) we get dy/dx = 2/4 = 1/2, so the line is
(y-1) = (x+2)/2
or y = x/2 + 2.

Answer 2

To find the equation of a tangent line, we need the point and slope.
We have the point (-2 , 1)
Now we need slope = derivaitve
we must use implicit differentiation

deriv of each side
2x + 8y y ' = 0
y ' = -2x / (8y) = - x / (4y) ====> plug in the point (-2,1)
y' = 2 / 4 = 1 / 2

Now use point-slope

y - 1 = (1/2) ( x + 2)
y = 1/2 x + 2
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Answer 3

The slope dy/dx is given at any point by:

2 x + 8 y dy/dx = 0.

Therefore dy/dx = - x/(4y) = 1/2 at (- 2, 1).

Thus, the tangent is y - 1 = 1/2 (x + 2), or

y = 1/2 x + 2.

CHECK: (i) This has the right slope, and (ii) For (x, y) = (- 2, 1), the equation correctly gives 1 = 1/2 (- 2) + 2 = - 1 + 2 = 1.

Live long and prosper.

Answer 4

first find the slope
2xdx + 8ydy = 0
2xdx = -8ydy
dy/dx = -x/4y
at (-2,1) dy/dy = 1/2

y=mx+b for tangent line

y=1/2x+b
at (-2,1) 1=1/2(-2) + b
b = 2
tangent line is y = 1/2x + 2

Answer 5

2x+8yy'=0
y'=-x/4y=1/2
y-1=1/2(x+2)
y=1/2x+2 or 2y-x-4=0