Question

Find the equations of the tangents to the curve
x = 6t^2 + 6,
y = 4t^3 + 2
that pass through the point (12, 6).

Please give any steps! Thank you!

Answer 1

Point (x, y) = (12, 6).

x = 6t² + 6

Substitute x value.

12 = 6t² + 6

6t² = 12 - 6

6t² = 6

t² = 6/6

t² = 1

t =  √1

t = ±1

y = 4t³ + 2

Substitute y value.

6 = 4t³ + 2

4t³ = 6 - 2

4t³ = 4

t³ = 4/4

t³ = 1

t = cube root of 1 = 1

Comparing 2 cases we conclue that t = 1.

x = 6t² + 6

dx/dt = d/dt(6t² + 6) = 12t = 12(1) = 12.

y = 4t³ + 2

dy/dt = d/dt(4t³ + 2) = 12t² = 12(1)² = 12.

Therefore slope of line m = dy/dx = 12/12 = 1.

Equation of the line in slope intercept form passing through (12, 6) is y - y1 = m(x - x1)

y - 6 = 1(x - 12)

y - 6 = 1(x) - 1(12)

y - 6 = x - 12

y = x - 12 + 6

y = x - 6 is the required tangent equation.