1. ## AS exam question

Another question from me.

The curve C has equation $y=x^2(x-6) + \frac{4}{x}, x>0$.

The Points P and Q lie on C and have x-coordinates 1 and 2 respectively.

a) Show that the length of PQ is $\sqrt 170$.
b) Show that the tangents to C at P and Q are parallel.
c) Find an equation for the noram to C at p giving your answer in the form $ax +by+c = 0$

No idea where to start with this equation the $+ \frac{4}{x}$ on the end really confuses me too, haven't seen that before.

2. Originally Posted by Ifailatmaths
Another question from me.

The curve C has equation $y=x^2(x-6) + \frac{4}{x}, x>0$.

The Points P and Q lie on C and have x-coordinates 1 and 2 respectively.

a) Show that the length of PQ is $\sqrt 170$.
b) Show that the tangents to C at P and Q are parallel.
c) Find an equation for the noram to C at p giving your answer in the form $ax +by+c = 0$

No idea where to start with this equation the $+ \frac{4}{x}$ on the end really confuses me too, haven't seen that before.
(a) Substitute the given values of x to get the y-coordinates of P and Q. Then use the distance formula.

(b) Find dy/dx and calculate its value at x = 1 and x = 2. What do you notice and therefore conclude ...? (To get the derivative, you should note that $\frac{4}{x} = 4 x^{-1}$).

(c) From (b) you know the gradient of the tangent. The gradient of the normal is -1/(gradient of tangent). From (a) you have a known point on the tangent. Use these two things to get the equation of the tangent as you would any line.