Question

The diagram shows the curve with equation $$y = 4x^{\frac{1}{2}}$$

(i) The straight line with equation $$y = x + 3$$ intersects the curve at points A and B. Find the length
of AB.

(ii) The tangent to the curve at a point T is parallel to AB. Find the coordinates of T.

(iii) Find the coordinates of the point of intersection of the normal to the curve at T with the line AB.

10(i)

$$4x^{\frac{1}{2}}=x+3\rightarrow (x^{\frac{1}{2}})^{2}-4x^{\frac{1}{2}}+3=0$$ OR  $$16x=x^{2}+6x+9$$

$$x^{\frac{1}{2}}=1$$  or 3

$$x^{2}-10x+9(=0)$$x=1or 9

y=4 or 12

AB=$$(9-1)^{2}+(12-4)^{2}$$

AB=$$\sqrt{128} or8\sqrt{2} oe or 11.3$$

10(ii)

$$\frac{dy}{dx}=2x^{\frac{-1}{2}}$$

$$2x^{\frac{-1}{2}}=1$$

(4,8)

10(iii)

Equation of normal is y $$-8=-1(x-4)$$

Eliminate $$y(or x)\rightarrow -x+12=x+3 or y-3=12-y$$

$$(4^{\frac{1}{2}},7^{\frac{1}{2}})$$

### Question

A diameter of a circle $$C_1$$ has end-points at (-3,-5) and (7,3).
(a) Find an equation of the circle $$C_1$$.

The circle $$C_1$$ is translated by $$\binom{8}{4}$$ to give circle $$C_2$$, as shown in the diagram.
(b) Find an equation of the circle $$C_2$$.
The two circles intersect at points R and S.
(c) Show that the equation of the line RS is y=-2x+13.
(d) Hence show that the x-coordinates of R and S satisfy the equation $$5x^2-60x+159=0$$.

Ans:

(a) Centre = (2, -1)
$$r^2=[2-(-3)]^2+[-1-(-5)]^2$$ or $$[2-7]^2+[-1-3]^2$$ OE
$$(x-2)^2+(y+1)^2=41$$
(b) Centre = their(2,-1)+$$\binom{8}{4}$$=(10,3)
$$(x-10)^2+(y-3)^2=their41$$
(c) Gradient m of line joining centres $$=\frac{4}{8}OE$$
Attempt to find mid-point of line.
Equation of RS is y-1=-2(x-6)
y=-2x+13
Alternative method for question 12(c)
$$(x-2)^2+(y+1)^2-41=(x-10)^2+(y-3)^2-41OE$$
$$x^2-4x+4+y^2+2y+1=x^2-20x+100+y^2-6y+9 OE$$
16x+8y=104
y=-2x+13
(d) $$(x-10)^2(-2x+13-3)^2=41$$
$$x^2-20x+100+4x^2-40x+100=41 \rightarrow 5x^2-60x+159=0$$