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.
Answer/Explanation
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 \(5x^2-60x+159=0\).
Answer/Explanation
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) \((x-10)^2(-2x+13-3)^2=41\)
\(x^2-20x+100+4x^2-40x+100=41 \rightarrow 5x^2-60x+159=0\)