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.
▶️Answer/Explanation
(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 \(y=7\sqrt{x}\)
▶️Answer/Explanation
(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.
Answer/Explanation
(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.
Answer/Explanation
(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.
Answer/Explanation
(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.
Answer/Explanation
(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]
Answer/Explanation
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.
Answer/Explanation
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.
Answer/Explanation
Ans:
x2 + 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]
Answer/Explanation
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.
Answer/Explanation
(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.
Answer/Explanation
(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