IB DP Further Mathematics -2.5 Finding equations of loci HL Paper 1

 

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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.

[4]
a.

Show that this equation represents a circle and state its radius and the coordinates of its centre.

[4]
b.
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]

a.

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]

b.

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 \(\theta \).

[3]
a.

As the wheel rolls, the point A traces out a curve. Show that the gradient of this curve is \(\cot \left( {\frac{1}{2}\theta } \right)\).

[6]
b.

Find the equation of the tangent to the curve when \(\theta  = \frac{\pi }{3}\).

[3]
c.
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

a.

\(\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = r – r\cos \theta \)     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

b.

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

c.

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]

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