**Question 15**:

Find the coordinates of the point on the curve,

*y*= (4*x*– 5)^{2}such that the gradient of the normal to the curve is $\frac{1}{8}$ .

*Solution:**y*= (4

*x*– 5)

^{2}

$\frac{dy}{dx}$
= 2(4

*x*– 5).4 = 32*x*– 40
Given the gradient of the normal is 1/8, therefore the gradient of the tangent is –8.

$\frac{dy}{dx}$
= –8

32

*x*– 40 = –8
32

*x*= 32*x*= 1

*y*= (4(1) – 5)

^{2}= 1

**Hence, the coordinates of the point on the curve,**

*y*= (4*x*– 5)^{2}is (1, 1).**Question 16**:

A curve has a gradient function of

*kx*^{2}– 7*x*, where*k*is a constant. The tangent to the curve at the point (1, 4) is parallel to the straight line*y*+ 2*x*–1 = 0. Find the value of*k*.

*Solution:*
Gradient function of

*kx*^{2}– 7*x*is*parallel to the straight line**y*+ 2*x*–1 = 0
$\frac{dy}{dx}$
=

*kx*^{2}– 7*x*

*y*+ 2

*x*–1 = 0,

*y*= –2

*x*+ 1, gradient of the straight line = –2

Therefore

*kx*^{2}– 7*x*= –2
At the point (1, 4),

*k*(1)

^{2}– 7(1) = –2

*k*– 7 = –2

*k*= 5**Question 17**:

In the diagram above, the
straight line

*PR*is normal to the curve $y=\frac{{x}^{2}}{2}+1$ at*Q*. Find the value of*k*.

*Solution:*$$\begin{array}{l}y=\frac{{x}^{2}}{2}+1\\ \frac{dy}{dx}=x\\ \\ \text{Atpoint}Q,\text{}x\text{-coordinate}=2,\\ \text{Gradientofthecurve,}\frac{dy}{dx}=2\\ \text{Hence,gradientofnormaltothecurve,}PR=-\frac{1}{2}\\ \frac{3-0}{2-k}=-\frac{1}{2}\\ 6=-2+k\\ k=8\end{array}$$

**Question 18**:

The normal to the curve

*y*=

*x*

^{2}+ 3

*x*at the point

*P*is parallel to the straight line

*y*= –

*x*+ 12. Find the equation of the normal to the curve at the point

*P.*

*Solution:*
Given normal to the curve at
point

*P*is parallel to the straight line*y*= –*x*+ 12.
Hence, gradient of normal to
the curve = –1.

As a result, gradient of
tangent to the curve = 1

*y*=

*x*

^{2}+ 3

*x*

*$\frac{dy}{dx}$*= 2

*x*+ 3

2

*x*+ 3 = 1
2

*x*= –2*x*= –1

*y*= (–1)

^{2}+ 3(–1)

*y*= –2

Point

*P*= (–1, –2).
Equation of the normal to the
curve at point

*P*is,*y*– (–2) = –1 (

*x*– (–1))

*y*+ 2 = –

*x*– 1

*y***= –**

*x*– 3