The Curve C has the equation 2x^2-11+13. The point Q lies on C such that the gradient of the normal to C at Q is -1/9. Find the x-co-ordinate of Q

The first ste here is the find the general equation for the gradient tangential to the curve. This is done by differentiation of the equation to give 4x-11=dy/dx. dy/dx is the gradient. Now we are given the gradient of the normal. As M_t*M_n=-1 we can find that the tangential gradient is 9. plugging this into the equation we can see that 4x-11=9. rearrage to find x so x=20/4 so x=5