Find the equations of the two tangentlines to the parabola y = x^2 - x that pass through the point (0,-6)

Asked on by sapun

1 Answer | Add Yours

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Lets consider the x coordinate of the tangent point be 'a'.

Then y cordinate will be `(a^2-a)`

The gradient of a line between two points is given by;

m = (difference in y coordinates)/(difference in x coordinates)


If the gradient of tangent line is m;

`m = (a^2-1-(-6))/(a-0) = (a^2-a+6)/a`


If the tangent line is y = mx+c we can write it as;

`y = (a^2-a+6)/a*x+c`

This line goes through (0,-6)

`-6 = (a^2-a+6)/a*0+c`

  `c = -6`


`y = (a^2-a+6)/a*x-6`


For any curve the gradient of the tangent is given by the first derivative of the function of curve.

`y' = 2x-1`

Gradient of the curve at point `(a,a^2-a)` ;

`m = 2a-1`


`2a-1 = (a^2-a+6)/a`

`2a^2-a = a^2-a+6`

     `a^2 = 6`

         `a = +-sqrt6`


`y = (6-sqrt6+6)/sqrt6*x-6 = (12-sqrt6)/sqrt6*x-6`

`y = (6-(-sqrt6)+6)/(-sqrt6)*x-6 = -(12+sqrt6)/sqrt6*x-6`


Equations of the tangent lines will be;

`y = (12-sqrt6)/sqrt6*x-6`

`y = -(12+sqrt6)/sqrt6*x-6`




We’ve answered 319,864 questions. We can answer yours, too.

Ask a question