As Aqa core 1 Maths question

1.) The circle with centre C (5,8) touches the y-axis.

The points P and Q lie on the circle, and the mid-point of PQ is M (7,12).
(i) Show that the lengths of CM is n√5 where n is an integer.
(ii) Hence find the area of triangle PCQ.

2.)The gradient dy/dx, of a curve C at the point (x,y) is given by
dy/dx= 20x-6x²-16
(ai) Show that y is increasing when 3x²-10x+8<0.
(aii) Solve the inequality 3x²-10x+8<0.

The curve C passes through the point P (2,3).
(bi.) Verify that the tangent to the curve at P is parallel to the x-axis.
(ii) The point Q (3,-1), also lies on the curve. The normal to the curve at Q and the tangent to the curve at P intersect at the point R. Find the co-ordinates of R.

Hello 'me'.

1(i)
CM use Pythagoras
CM = sqrt( (7-5)^2 + ( 12-8)^2)
CM = sqrt( 4 + 16)
CM = sqrt(20)
CM = 2sqrt(5) n = 2

(ii)
Since the circle centre is (5,8) and it touches the y-axis, then the distance from the centre to the y-axis is '5'. ( The difference in 'x').
Since P & Q lie on the circle then then the distance from C is also '5' ( Radii).
CP = CQ = 5
Since M is the mid point of PQ then it intercepts PQ at right angles. and its ddistance is 2sqrt(5)
We now have a triangle CPM and applying Pythagoras to find PM
5^2 = PM^2 + (2sqrt(5))^2
25 = PM^2 + 20
PM^2 = 25 - 20 = 5
PM = sqrt(5)

The area of a triangler is A = 0.5bh.
b= sqrt(5)
h = 2sqrt(5)
A(PCM) = 0.5 x sqrt(5) x 2sqrt(5)
A = = 0.5 x 2 x 5
A = 5
Since the other triangle QCM is congruent to PCM then the area is doubled to 10 .
So area of PCQ = 10

2a(i)
Doubly differentiate
d2y/dx^2 = 6x - 10 < 0
6x < 10
x < 10/6 = 5/3
So when x < 5/3 then y is increasing.

(ii)
3x^2 - 10x + 8 < 0
Factorise
(3x - 4)(x - 2) < 0
3x - 4 > 0
3x > 4
x > 4/3