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

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

(1,1),(4,4)

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

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

x=-2,y=16

(ii)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}\0 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)$$y^{2}=3y\Rightarrow y(y-3)=0\Rightarrow y=3$$  or  0

$$x=\frac{1}{2}$$ or 5 $$\Rightarrow a=5$$

(ii)$$\left [ \frac{\left ( 2x-1 \right )^{\frac{3}{2}}}{\frac{3}{2}} \right ]\div 2$$

$$\left [ \frac{2}{3}\times \frac{x^{2}}{2}-\frac{x}{3} \right ]$$

$$\left [ \frac{27}{9}-0 \right ]$$            $$\left [ \frac{25}{3}-\frac{5}{3}-\left ( \frac{1}{12}-\frac{1}{6} \right ) \right ]$$

$$\frac{9}{4}$$

Question

A curve has equation $$y=x^{2}-x+3$$  and a line has equation y = 3x + a, where a is a constant.
(i) Show that the x-coordinates of the points of intersection of the line and the curve are given by the equation $$x^{2}-4x+(3+a)=0$$.
(ii) For the case where the line intersects the curve at two points, it is given that the x-coordinate of one of the points of intersection is −1. Find the x-coordinate of the other point of intersection.
(iii) For the case where the line is a tangent to the curve at a point P, find the value of a and the coordinates of P.

(i)$$x^{2}-x+3=3x+a\rightarrow x^{2}-4x+(3-a)=0$$

(ii)$$5+(3-a)=0\rightarrow a=8$$

$$x^{2}-4x-5=0\rightarrow x=5$$

(iii) $$16-4\left ( 3-a \right )=0$$  (applying $$b^{2}-4ac=0$$)

a=-1

$$(x-2)^{2}=0\rightarrow x=2,y=5$$

#### Question.

The equation of a curve is y = 2 cos x.
(i) Sketch the graph of y = 2 cos x for −π ≤ x ≤ π, stating the coordinates of the point of intersection
with the y-axis.
Points P and Q lie on the curve and have x-coordinates of $$\frac{\pi}{3}$$ and $$\pi$$ respectively.

(ii) Find the length of PQ correct to 1 decimal place.

The line through P and Q meets the x-axis at H (h, 0) and the y-axis at K(0, k).

(iii) Show that $$h =\frac{5\pi}{9}$$ and find the value of k.

(i)

(ii)$$P(\frac{\pi }{3},1)Q\left ( \pi ,-2 \right )$$

$$\rightarrow PQ^{2}=\left ( \frac{2\pi }{3} \right )^{2}+3^{2}\rightarrow PQ=3.7$$

(iii)Equation of PQ $$y-1=-\frac{9}{2\pi }\left ( x-\frac{\pi }{3} \right )$$

If y=0$$\rightarrow h=\frac{5\pi }{9}$$

If x=0$$\rightarrow k=\frac{5}{2}$$

Question

The line 4y = x + c, where c is a constant, is a tangent to the curve $$y^{2}=x+3$$

(i) Find the value of c.

(ii) Find the coordinates of P.

(i) Eliminates x or y→$$y^{2}-4y+c-3=0 or x^{2}+(2c-16)x+c^{2}-48=0 uses b^{2}= 4ac\rightarrow 4c-28=0$$ c=7

Alternative method for question 2(i)$$\frac{dy}{dx}=\frac{1}{\sqrt[2]{(x+3)}}=\frac{1}{4}$$

Solving c = 7

(ii) Uses c = 7,$$y^{2}$$− 4y + 4 = 0 (1, 2)

### Question

Find the set of values of m for which the line with equation y = mx − 3 and the curve with equation
y = 2x2 + 5 do not meet.                                                                                                                                           [3]

Ans

1 $$2x^{2}+5=mx-3\rightarrow 2x^{2}-mx+8(=0)$$

m2 − 64
− 8 <m <8

### Question

A line has equation y = 3x + k and a curve has equation y = x2 + kx + 6, where k is a constant.
Find the set of values of k for which the line and curve have two distinct points of intersection.

Ans:

$$x^2+kx+6=3x+k$$ leading to $$x^2+x(k-3)+(6-k)[=0]$$

$$(k-3)^2-4(6-k)[>0]$$

$$(k^2-2k-15[>0]$$

k<-3, k>5

### Question

A curve has equation y = x2 + 2cx + 4 and a straight line has equation y = 4x + c, where c is a constant.
Find the set of values of c for which the curve and line intersect at two distinct points.

Ans:

x+ 2cx + 4 = 4x + c  leading to x2 + 2cx – 4x + 4 – c [=0]

b2 – 4ac = (2c – 4)2 – 4(4-c)

[4c2 – 16c + 16 – 16 + 4c=] 4c2 – 12

b2 – 4ac > 0 leading to (4) c (c-3) > 0

c<0, c>3

### Question

The equation of a circle is x2 + y2 − 4x + 6y − 77 = 0.
(a) Find the x-coordinates of the points A and B where the circle intersects the x-axis.    [2]

(b) Find the point of intersection of the tangents to the circle at A and B.                            [6]

Ans

10 (a) When y = 0  x2 – 4x – 77 = 0 [⇒ (x+7 )(x-11 ) = 0 or (x – 2)2 = 81 ]

So x-coordinates are  −7 and 11

10 (b) Centre of circle C is (2,−3)

Gradient of AC is  $$-\frac{1}{3}$$  or Gradient of BC is $$-\frac{1}{3}$$

Gradient of tangent at A is 3 or Gradient of tangent at B is −3

Equations of tangents are y = 3 x + 21, y = −3x + 33

Meet when 3x + 21 = −3x + 33

Coordinates of point of intersection (2, 27)

Alternative method for Question 10(b)

Implicit differentiation:  $$2y\frac{dy}{dx}\ seen$$

$$2x-4+2y\frac{dy}{dx}+6\frac{dy}{dx}=0$$

Gradient of tangent at A is 3 or Gradient of tangent at B is  −3

Equations of tangents are y = 3x + 21, y = −3x + 33

Meet when 3x + 21 = −3x + 33

Coordinates of point of intersection (2, 27)

### Question.

The equation of a line is $$y =mx + c$$, where m and c are constants, and the equation of a curve is $$xy = 16$$.
(a) Given that the line is a tangent to the curve, express m in terms of c.

(b) Given instead that $$m =−4$$, find the set of values of c for which the line intersects the curve at two distinct points.

(a)  $$x(mx + c) = 16\rightarrow mx2 + cx−16 =0$$

Use of $$b² − 4ac = c² + 64m$$

Sets to $$0 →m =\frac{-c^{2}}{64}$$

(b) $$x(− 4x + c) = 16$$
Use of $$b² − 4ac→ c² − 256$$

$$c > 16$$ and $$c <−16$$

### Question.

The diagram shows part of the curve $$y =\frac{8}{x+2}$$ and the line $$2y + x = 8$$, intersecting at points A and B.  The point C lies on the curve and the tangent to the curve at C is parallel to AB.
(a) Find, by calculation, the coordinates of A, B and C.

(b) Find the volume generated when the shaded region, bounded by the curve and the line, is rotated through $$360^{\circ}$$ about the x-axis.

(a) Simultaneous equations $$\frac{8}{x+2}=4=\frac{1}{2}x$$

x = 0 or x = 6→ A (0, 4) and B (6, 1)

At $$C=\frac{-8}{(x+2)^{2}}=-\frac{1}{2}\rightarrow C(2,2)$$

(B1 for the differentiation. M1 for equating and solving)

(b) Volume under line $$\pi \int (-\frac{1}{2}x+4)^{2}dx=\left [ \frac{x^{3}}{12}-2x^{2}+16x \right ]=(42\pi )$$

(M1 for volume formula. A2,1 for integration)

Volume under curve $$=\pi \int \left ( \frac{8}{x+2} \right )^{2}dx=\pi \left [ \frac{-64}{x+2} \right ]=(24\pi )$$

Subtracts and uses 0 to 6→ 18π

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