<p>A curve \(C\) has parametric equations \(x = 3t^2\), \(y = 2t^3\), for&nbsp;\(t &lt; 0\).</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Sketch \(C\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Points \(P\) and \(Q\) have parameters \(p\) and \(q\) respectively. Given that&nbsp;\(q = 2p\), find the cartesian equation of the locus of the mid-point of \(PQ\) as&nbsp;\(p\) varies.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>A curve \(L\) has equation \(x = y^2 + 1\).</p>
<ol class="i-list" start="4"><li><div class="list-item-paragraph">It is given that the curves \(C\) and \(L\) touch at the point \(M\). Show that&nbsp;\(4t^6 - 3t^2 + 1 = 0\) at \(M\).<br>Hence, or otherwise, find the exact coordinates of \(M\).<span class="inline-marks">[6]</span></div></li></ol>

A curve \(C\) has parametric equations \(x = 3t^2\), \(y = 2t^3\), for \(t < 0\).

Points \(P\) and \(Q\) have parameters \(p\) and \(q\) respectively. Given that \(q = 2p\), find the cartesian equation of the locus of the mid-point of \(PQ\) as \(p\) varies.

A curve \(L\) has equation \(x = y^2 + 1\).

It is given that the curves \(C\) and \(L\) touch at the point \(M\). Show that \(4t^6 - 3t^2 + 1 = 0\) at \(M\).\nHence, or otherwise, find the exact coordinates of \(M\).

It is given that the curves \(C\) and \(L\) touch at the point \(M\). Show that \(4t^6 - 3t^2 + 1 = 0\) at \(M\).
Hence, or otherwise, find the exact coordinates of \(M\).

Parametric Equations

Graphing Techniques

International Baccalaureate Higher Level Mathematics

IB HL Math

N2013 P1 Q11

Since \(t < 0\): \(x = 3t^2 > 0\) and \(y = 2t^3 < 0\).

When \(t \to 0^-\): \((x, y) \to (0, 0)\) (open circle).

Curve lies entirely in the fourth quadrant, starting from near the origin and going downward to the right.

\(P(3p^2, 2p^3)\), \(Q(3(2p)^2, 2(2p)^3) = Q(12p^2, 16p^3)\)

Midpoint of \(PQ\), \(N = \left(\dfrac{3p^2 + 12p^2}{2}, \dfrac{2p^3 + 16p^3}{2}\right) = \left(\dfrac{15}{2}p^2, 9p^3\right)\)

\(x = \dfrac{15}{2}p^2\), \(y = 9p^3\)

From \(y = 9p^3\): \(p = \left(\dfrac{y}{9}\right)^{\frac{1}{3}}\)

Sub into \(x = \dfrac{15}{2}p^2\):

\(x = \dfrac{15}{2}\left(\dfrac{y}{9}\right)^{\frac{2}{3}}\), \(y < 0\)

Sub \(x = 3t^2\), \(y = 2t^3\) into \(L: x = y^2 + 1\):

\(3t^2 = (2t^3)^2 + 1\)

\(3t^2 = 4t^6 + 1\)

\(4t^6 - 3t^2 + 1 = 0\) (shown)

Let \(v = t^2\): \(4v^3 - 3v + 1 = 0\)

\(v = -1\) is a root \(\Rightarrow (v+1)\) is a factor.

By long division: \(4v^3 - 3v + 1 = (v+1)(4v^2 - 4v + 1) = (v+1)(2v-1)^2\)

\((2v-1)^2(v+1) = 0\)

\(v = \dfrac{1}{2}\) (repeated) or \(v = -1\)

\(t^2 = \dfrac{1}{2}\) or \(t^2 = -1\) (rejected, \(t \in \mathbb{R}\))

\(t = \dfrac{1}{\sqrt{2}}\) or \(t = -\dfrac{1}{\sqrt{2}}\)

Since \(t < 0\): \(t = -\dfrac{1}{\sqrt{2}}\)

\(x = 3\left(-\dfrac{1}{\sqrt{2}}\right)^2 = 3\left(\dfrac{1}{2}\right) = \dfrac{3}{2}\)

\(y = 2\left(-\dfrac{1}{\sqrt{2}}\right)^3 = 2\left(-\dfrac{1}{2\sqrt{2}}\right) = -\dfrac{1}{\sqrt{2}}\)

\(\therefore M\left(\dfrac{3}{2}, -\dfrac{1}{\sqrt{2}}\right)\)

N2013 P1 Q11

N2013 P1 Q11

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