[CALCULUS] Find the Equation of The Tangent Line through the point (1,2) on the curve x2+4y2-4x-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

x² +4y² -4x -8y + 3 = 0

Differentiate implicitly with respect to x: 2x + 8y(dy/dx) - 4 - 8(dy/dx) = 0

dy/dx = (-2x + 4) / (8y - 8)

Slope of tangent line = dy/dx evaluated at the point (1,2) = 1/4

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

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