# CIE A level -Pure Mathematics 1 : Topic : 1.3 Coordinate geometry: graph and its associated algebraic equation : Exam Questions Paper 1

### Question

A curve has equation $$y=x^{2}-4x+4$$ and a line has equation y=mx  ,where m is a constant.

(i)For the case where m=1,the curve and the line intersect at the points A and B.Find the coordinates of mid-point of AB.

(ii)Find the non-zero value of m for which the line is a tangent to the curve ,and find the coordinates of the point where the tangent touches the curve.

(i)$$x^{2}-4x+4=x\Rightarrow x^{2}-5x+4=0$$

$$\left ( x-1 \right )\left ( x-4 \right )=0$$

So, the points of intersection A and B are $\left(1,1\right),$(4,4).

By using midpoint formula,$$M\left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right)$$

Mid-point=$$\left ( 2\tfrac{1}{2},2\tfrac{1}{2} \right )$$

(ii)$$x^{2}-\left ( 4+m \right )x+4=0\rightarrow \left ( 4+m \right )^{2}-4(4)=0$$

$$4+m=\pm 4$$ or $$m\left ( 8+m \right )=0$$

m=-8

$$x^{2}+4x+4=0$$

So, the coordinates of the point where the tangent touches the curve are $\left(-2,16\right).$

OR

2x-4=m

$$x^{2}-4x+4=\left ( 2x-4 \right )x$$

x=-2 (ignore +2)

m=-8(ignore 0)

y=16

Question

The diagram shows the curve  and the line $$y=7\sqrt{x}$$ and the line y=6c+k ,where k is a constant. The curve and the line intersect at the points A and B

(i)For the case where k=2,find the x-coordinates of A and B.

(ii)Find the value of k for which y=6x+k is a tangent to the curve $$y=7\sqrt{x}$$

(i)$$6x+2=7\sqrt{x}\Rightarrow 6\left ( \sqrt{x} \right )^{2}-7\sqrt{x}+2=0$$

$$\left ( 3\sqrt{x} -2\right )\left ( 2\sqrt{x} -1\right )=0$$

$$\sqrt{x}=\frac{2}{3}$$ or $$\frac{1}{2}$$

$$x=\frac{4}{9}$$ or $$\frac{1}{4}\0 (or 0.444,0.25) OR \(\left ( 6x+2 \right )^{2}=49x\rightarrow 36x^{2}-25x+4=0$$

$$\left ( 9x-4 \right )\left ( 4x-1 \right )=0$$

$$x=\frac{4}{9}$$ or $$\frac{1}{4}\0 (or 0.444,0.25) (ii)\(7^{2}-4\times6\times k=0$$

$$k=\frac{49}{24}$$ or 2.04

OR

$$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 7x^{\frac{1}{2}} \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left ( 6x+k \right )\rightarrow \frac{7}{2}x^{-\frac{1}{2}}=6$$

$$x=\frac{49}{144},y=\frac{49}{12}\rightarrow k=\frac{49}{24}$$ or 2.04

### Question

The diagram show  the curve $$y^{2}=2x-1$$  and the straight line 3y=2x-1.The curve and thew straight line intersect at $$x=\frac{1}{2}$$ and x=a and ,where a is a constant.

(i)Show that a=5.

(ii)Find,showing all necessary working ,the area of the shaded region.

(i) To find the value of $$a$$, we know that the curve $$y^2 = 2x – 1$$ intersects the line $$3y = 2x – 1$$ at $$x = \frac{1}{2}$$ and $$x = a$$.
First, substitute $$x = \frac{1}{2}$$ into the equation of the line to find the corresponding $$y$$:
$$3y = 2\left(\frac{1}{2}\right) – 1$$
$$3y = 0 \implies y = 0$$
So, at $$x = \frac{1}{2}$$, the point of intersection is $$\left(\frac{1}{2}, 0$$\).
Now, substitute $$x = a$$ into both equations:
$$y^2 = 2a – 1$$
$$3y = 2a – 1$$
$$2a – 1 = y^2$$
Now, substitute $$x = \frac{1}{2}$$ and $$y = 0$$:
$$2\left(\frac{1}{2}\right) – 1 = 0^2$$
This is true, confirming that $$x = \frac{1}{2}$$. Therefore, $$a = 5$$.
(ii) Now, we need to find the area of the shaded region. The shaded region is the area between the curve $$y^2 = 2x – 1$$ and the line $$3y = 2x – 1$$ from $$x = \frac{1}{2}$$ to $$x = 5$$.
The area $$A$$ is given by the integral:
$$A = \int_{\frac{1}{2}}^5 \left(3y – \frac{1}{2}\right) \,dx$$
To find $$y$$ in terms of $$x$$, we use the equation $$3y = 2x – 1$$, so $$y = \frac{2x – 1}{3}$$.
Substitute this into the integral:
$$A = \int_{\frac{1}{2}}^5 \left(3 \cdot \frac{2x – 1}{3} – \frac{1}{2}\right) \dx$$
$$\left [ \frac{\left ( 2x-1 \right )^{\frac{3}{2}}}{\frac{3}{2}} \right ]\div 2$$
$$\left [ \frac{2}{3}\times \frac{x^{2}}{2}-\frac{x}{3} \right ]$$
$$\left [ \frac{27}{9}-0 \right ]$$
$$\left [ \frac{25}{3}-\frac{5}{3}-\left ( \frac{1}{12}-\frac{1}{6} \right ) \right ]$$
$$\frac{9}{4}$$

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