Download IITian Academy App for accessing Online Mock Tests and More..
Question
The point \({\rm{P}}(x,y)\) moves in such a way that its distance from the point (\(1\) , \(0\)) is three times its distance from the point (\( -1\) , \(0\)) .
Find the equation of the locus of P.
Show that this equation represents a circle and state its radius and the coordinates of its centre.
Answer/Explanation
Markscheme
We are given that
\(\sqrt {{{(x – 1)}^2} + {y^2}} = 3\sqrt {{{(x + 1)}^2} + {y^2}} \) M1A1
\({x^2} – 2x + 1 + {y^2} = 9({x^2} + 2x + 1 + {y^2})\) A1
\(8{x^2} + 8{y^2} + 20x + 8 = 0\) A1
[4 marks]
Rewrite the equation in the form
\({\left( {x + \frac{5}{4}} \right)^2} + {y^2} = – 1 + \frac{{25}}{{16}} = \frac{9}{{16}}\) M1A1
This represents a circle with radius \( = \frac{3}{4}\) ; centre \(\left( { – \frac{5}{4},0} \right)\) A1A1
Note: Allow FT from the line above.
[4 marks]
Question
The parabola \(P\) has equation \({y^2} = 4ax\). The distinct points \({\text{U}}\left( {a{u^2},{\text{ }}2au} \right)\) and \({\text{V}}\left( {a{v^2},{\text{ }}2av} \right)\) lie on \(P\), where \(u,{\text{ }}v \ne 0\). Given that \({\rm{U\hat OV}}\) is a right angle, where \({\text{O}}\) denotes the origin,
(a) show that \(v = – \frac{4}{\mu }\);
(b) find expressions for the coordinates of \({\text{W}}\), the midpoint of \([{\text{UV}}]\), in terms of \(a\) and \(u\);
(c) show that the locus of \({\text{W}}\), as \(u\) varies, is the parabola \({P’}\) with equation \({y^2} = 2ax – 8{a^2}\);
(d) determine the coordinates of the vertex of \({P’}\).
Answer/Explanation
Markscheme
(a) gradient of \({\text{OU}} = \frac{{2au}}{{a{u^2}}} = \frac{2}{u}\) A1
gradient of \({\text{OV}} = \frac{{2av}}{{a{v^2}}} = \frac{2}{v}\) A1
since the lines are perpendicular,
\(\frac{2}{u} \times \frac{2}{v} = – 1\) M1
so \(v = – \frac{4}{u}\) AG
[3 marks]
(b) coordinates of \({\text{W}}\) are \(\left( {\frac{{a({u^2} + {v^2})}}{2},{\text{ }}\frac{{2a(u + v)}}{2}} \right)\) M1
\( = \left( {\frac{a}{2}\left( {{u^2} + \frac{{16}}{{{u^2}}}} \right),{\text{ }}a\left( {u – \frac{4}{u}} \right)} \right)\) A1
[2 marks]
(c) putting
\(x = \frac{a}{2}\left( {{u^2} + \frac{{16}}{{{u^2}}}} \right);{\text{ }}y = a\left( {u – \frac{4}{u}} \right)\) M1
it follows that
\({y^2} = {a^2}\left( {{u^2} + \frac{{16}}{{{u^2}}} – 8} \right)\) A1
\( = 2ax – 8{a^2}\) AG
Note: Accept verification.
[2 marks]
(d) since \({y^2} = 2a(x – 4a)\) (M1)
the vertex is at \((4a,{\text{ }}0)\) A1
[2 marks]
Question
A wheel of radius \(r\) rolls, without slipping, along a straight path with the plane of the wheel remaining vertical. A point \({\text{A}}\) on the circumference of the wheel is initially at \({\text{O}}\). When the wheel is rolled, the radius rotates through an angle of \(\theta \) and the point of contact is now at \({\text{B}}\), where the length of the arc \({\text{AB}}\) is equal to the distance \({\text{OB}}\). This is shown in the following diagram.
Find the coordinates of \({\text{A}}\) in terms of \(r\) and \(\theta \).
As the wheel rolls, the point A traces out a curve. Show that the gradient of this curve is \(\cot \left( {\frac{1}{2}\theta } \right)\).
Find the equation of the tangent to the curve when \(\theta = \frac{\pi }{3}\).
Answer/Explanation
Markscheme
\({\text{OX}} = {\text{OB}} – {\text{XB}} = r\theta – r\sin \theta = x\) (M1)A1
\({\text{OY}} = {\text{ZB}} – {\text{ZC}} = r – r\cos \theta = y\) A1
\(\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = r – r\cos \theta \) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}\theta }} = r\sin \theta \) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{r\sin \theta }}{{r – r\cos \theta }}\) M1
\( = \frac{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\sin }^2}\frac{\theta }{2}}}\) M1A1A1
\( = \frac{{\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}}\)
\( = \cot \frac{\theta }{2}\) AG
when \(\theta = \frac{\pi }{3}\), gradient \( = \sqrt 3 \) A1
\(x = \frac{\pi }{3}r – r\frac{{\sqrt 3 }}{2},{\text{ }}y = r – \frac{r}{2} = \frac{r}{2}\) A1
\(y – \frac{r}{2} = \sqrt 3 \left( {x – \frac{\pi }{3}r + r\frac{{\sqrt 3 }}{2}} \right)\;\;\;{\text{or}}\;\;\;y = \sqrt 3 x + 2r – \frac{{\pi r}}{{\sqrt 3 }}\) A1
Question
A circle \({x^2} + {y^2} + dx + ey + c = 0\) and a straight line \(lx + my + n = 0\) intersect. Find the general equation of a circle which passes through the points of intersection, justifying your answer.
Answer/Explanation
Markscheme
METHOD 1
\({x^2} + {y^2} + dx + ey + c + \lambda (lx + my + n) = 0\) M1 A1
ie\(\;\;\;\)\({x^2} + {y^2} + x(d + \lambda l) + y(e + \lambda m) + c + \lambda n = 0\) A1
since \({x^2}\) and \({y^2}\) have the same coefficients and there is no \(xy\) term, this is a circle R1
we know the pair of points fit the equation. R1
hence this is the required equation.
METHOD 2
Let the general equation be
\({x^2} + {y^2} + ax + by + q = 0\) M1
The intersection with the given circle satisfies
\((a – d)x + (b – e)y + (q – c) = 0\) M1A1
This must be the same line as \(lx + my + n = 0\) R1
Therefore
\(a – d = \lambda l\;\;\;{\text{giving}}\;\;\;a = d + \lambda l\)
\(b – e = \lambda m\;\;\;{\text{giving}}\;\;\;b = e + \lambda m\) A1
\(q – c = \lambda n\;\;\;{\text{giving}}\;\;\;q = c + \lambda n\)
leading to the required general equation
Note: Award M1 to candidates who only attempt to find the points of intersection of the line and circle
Question
Given that the tangents at the points P and Q on the parabola \({y^2} = 4ax\) are perpendicular, find the locus of the midpoint of PQ.
Answer/Explanation
Markscheme
EITHER
attempt to differentiate (M1)
let \(y = 2at \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}t}} = 2a\) and \(x = a{t^2} \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}t}} = 2at\) A1
hence \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}t}} \times \frac{{{\text{d}}t}}{{{\text{d}}x}} = \frac{{2a}}{{2at}} = \frac{1}{t}\) A1
let P have coordinates \(\left( {at_1^2,\,2a{t_1}} \right)\) and Q have coordinates \(\left( {at_2^2,\,2a{t_2}} \right)\) (M1)
therefore gradient of tangent at P is \(\frac{1}{{{t_1}}}\) and gradient of tangent at Q is \(\frac{1}{{{t_2}}}\) A1
since these tangents are perpendicular \(\frac{1}{{{t_1}}} \times \frac{1}{{{t_2}}} = – 1 \Rightarrow {t_1}{t_2} = – 1\) A1
mid-point of PQ is \(\left( {\frac{{a\left( {t_1^2 + t_2^2} \right)}}{2},\,\,a\left( {{t_1} + {t_2}} \right)} \right)\) A1
\({y^2} = {a^2}\left( {t_1^2 + 2{t_1}{t_2} + t_2^2} \right)\) M1
\({y^2} = {a^2}\left( {\frac{{2x}}{a} – 2} \right)\,\,\,\left( { \Rightarrow {y^2} = 2ax – 2{a^2}} \right)\) A1
OR
attempt to differentiate (M1)
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4a\) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2a}}{y}\)
let coordinates of P be \(\left( {{x_1},\,{y_1}} \right)\) and the coordinates of Q be \(\left( {{x_2},\,{y_2}} \right)\) (M1)
coordinates of midpoint of PQ are \(\left( {\frac{{{x_1} + {x_2}}}{2},\,\,\frac{{{y_1} + {y_2}}}{2}} \right)\) M1
if the tangents are perpendicular \(\frac{{2a}}{{{y_1}}} \times \frac{{2a}}{{{y_2}}} = – 1\) A1
\( \Rightarrow {y_1}{y_2} = – 4{a^2}\)
\(y_1^2 + y_2^2 = 4a\left( {{x_1} + {x_2}} \right)\) A1
\(\frac{{y_1^2 + 2{y_1}{y_2} + y_2^2}}{4} = \frac{{4a\left( {{x_1} + {x_2}} \right) + 2{y_1}{y_2}}}{4}\) M1
\(\left( {\frac{{{y_1} + {y_2}}}{2}} \right) = 2a\frac{{{x_1} + {x_2}}}{2} – \frac{{8{a^2}}}{4}\) A1
hence equation of locus is \({y^2} = 2ax – 2{a^2}\) A1
[9 marks]