Question

The equation of the tangents of the curve x = t^2 + t - 2 and y = t^3 + 3t -4, parallel to the line 3x - y + 3 = 0;?

Answer 1

The parametric equations are x = t ^2 + t - 2 and y = t ^3 + 3t - 4.

dx / dt = 2t + 1

dy / dt = 3t ^2 + 3

dy / dx = (dy / dt) / (dx / dt)

          = (3t  ^2 + 3) / (2t + 1).

Here, dy / dx is slope of the tangent line.

The tangent line is parallel to 3x - y + 3 = 0.

So, parallel lines have same slopes.

Write the equation 3x - y + 3 = 0 in slope intercept form y = mx + c, where m is slope and c is y - intercept.

y = 3x + 3.

Compare the above equation with y = mx + c.

slope is 3.

Therefore, slope of the tangent line is is equals to 3.

⇒ (3t  ^2 + 3) / (2t + 1) = 3

3t ^2 + 3 = 3(2t + 1)

3t t^2 + 3 = 6t + 3

3t ^2 - 6t = 0

t ^2 - 2t = 0

t (t   - 2) = 0

⇒ t   = 0 and t  = 2.

• Substitute t  = 0 in x = t ^2 + t - 2 and y = t ^3 + 3t - 4.

x = 0^2 + 0 - 2 and y = 0^3 + 3(0) - 4

⇒ x = - 2 and y = - 4.

Therefore, the point is (- 2, - 4)

To find the tangent line equation, substitute the values of m = 3 and (x, y ) = (- 2, - 4)  in the slope intercept form of an equation.

y = mx + c.

- 4 = ( 3)( - 2) + c.

c = - 4 + 6

c = 2.

Substitute m = 3 and c = 2 in y = mx + c.

y = ( 3 )x + 2.

The tangent line equation is y = 3x + 2.

• Substitute t  = 2 in x = t ^2 + t - 2 and y = t ^3 + 3t - 4.

x = 2^2 + 2 - 2 and y = 2^3 + 3(2) - 4

⇒ x  = 4 and y  = 10.

Therefore, the point is (4, 10)

To find the tangent line equation, substitute the values of m = 3 and (x, y ) = (4, 10)  in the slope intercept form of an equation.

y = mx + c.

10 = ( 3)( 4) + c.

c = 10 - 12

c = - 2.

Substitute m = 3 and c  = - 2 in y = mx + c.

y = ( 3 )x + (- 2)

The tangent line equation is y = 3x - 2.