Find the Equation of The Tangent Line through the point (1,2) on the curve x^2+4y^2-4x-8y+3=0

Question

Find the Equation of The Tangent Line through the point (1,2) on the curve x²+4y²-4x-8y+3=0

Answer 1

Here we need to differentiate both variables as they are related rates.

x2 + 4y2 - 4x - 8y + 3 = 0

2x + 8yy' - 4 - 8y' = 0

8yy' - 8y' = 4 - 2x

8y'(y - 1) = 2(2 - x)

y' = [2(2 - x)] / [8(y - 1)] = slope of the tangent line

At the point (1,2) y' = [2(2 - (1))] / [8((2) - 1)] = 1/4

Now that we have the slope and a point we can get the equation of the line.

(y - 2) = (1/4)(x - 1)

y = (1/4)(x + 7)

Answer 2

Let find derivative implicitly: 2x + 8yy' - 4 - 8y' = 0; 4y'(y - 1) = 2 - x; y' = (2 - x)/[4(y - 1)].

Slope m of tangent line at point (1, 2) m = (2 - 1)/[4(2-1)] = 1/4

Equation of tangent line: y - 2 = 1/4(x - 1); y = 1/4x - 1/4 + 2; y = 1/4x + 7/4