Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Answer 1

Answer:

y = 2(x - 1/4)

Step-by-step explanation:

To find an equation of the tangent to a given curve, we need two vital information; one is the slope of the tangent (SOP) and the other is the point of tangency (POT).

to find the slope of tangent i will use the implicit differentiation to find the slope then apply the point given

so, first, the slope using implicit differentiation;

f(x) = $x^{2} +y^{2} =(3x^{2}+4y^{2} -x) ^{2}$

f'(x) = {tex}2x + 2y dy/dx = 2(3x^{2} + 4y^{2} - x)(6x + 8y dy/dx - 1){/tex}

applying the co-ordinates given; x = 0 and y = 1/4

SOP  2(0) + 2(1/4) dy/dx = 2[3(0)^2 + 4(1/4)^2 - 0][6(0) + 8(1/4) dy/dx - 1)

1/2 dy/dx = 2[4(1/16)][(8/4) dy/dx - 1]

1/2 dy/dx = 2[1/2][2 dy/dx - 1]

1/2 dy/dx = 2 dy/dx - 1

putting the dy/dx together

1 = 2 dy/dx - 1/2 dy/dx

1 = dy/dx (2 - 1/2)

make dy/dx subject

1 / 1/2 = dy/dx

therefore SOP = 2

POT

f(x) = (0)^2 + (1/4)^2 = [3(0)^2 + 4(1/4)^2 - 0]^2

1/16 = [4(1/16)]^2

1/16 = (1/4)^2

1/16 = 1/16

y = 0

using the slope formula

m = y - y2/ x - x2

2 = y - 0/ x - 1/4

y = 2(x - 1/4)