<p>The curve $C$ has parametric equations</p>
<p style="text-align: center">$x={{t}^{2}}+4t$, $y={{t}^{3}}+{{t}^{2}}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Sketch the curve for $-2\le t\le 1$.</span></div><p style="text-align: right;">[1]</p></li></ol>
<p>The tangent to the curve at point $P$ where $t=2$ is denoted by $l$.</p>
<ol class="i-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Find the cartesian equation of $l$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">The tangent $l$ meets $C$ again at the point $Q$. Use a non-calculator method to find the coordinates of $Q$.</span></div><p style="text-align: right;">[4]</p></li></ol>

$x={{t}^{2}}+4t$, $y={{t}^{3}}+{{t}^{2}}$.

The tangent to the curve at point $P$ where $t=2$ is denoted by $l$.

The tangent $l$ meets $C$ again at the point $Q$. Use a non-calculator method to find the coordinates of $Q$.

Parametric Equations

Graphing Techniques

International Baccalaureate Higher Level Mathematics

IB HL Math

N2009 P2 Q1

When $t = -2$: $x = 4 - 8 = -4$, $y = -8 + 4 = -4$. Point: $(-4, -4)$

When $t = 0$: $x = 0$, $y = 0$. Point: $(0, 0)$

When $t = 1$: $x = 1 + 4 = 5$, $y = 1 + 1 = 2$. Point: $(5, 2)$

[Sketch shows curve passing through $(-4, -4)$, $(0, 0)$, and $(5, 2)$]

$\frac{dx}{dt} = 2t + 4$, $\frac{dy}{dt} = 3t^2 + 2t$

$\frac{dy}{dx} = \frac{3t^2 + 2t}{2t + 4}$

At $t = 2$: $x = 4 + 8 = 12$, $y = 8 + 4 = 12$

$\left.\frac{dy}{dx}\right|_{t=2} = \frac{3(4) + 2(2)}{2(2) + 4} = \frac{16}{8} = 2$

Equation of tangent: $y - 12 = 2(x - 12)$

Substituting $x = t^2 + 4t$ and $y = t^3 + t^2$ into $y = 2x - 12$:

Since $t = 2$ corresponds to point $P$, $(t - 2)$ is a factor.

By polynomial division: $t^3 - t^2 - 8t + 12 = (t - 2)(t^2 + t - 6)$

$t = 2$ (rejected, this is $P$) or $t = -3$

When $t = -3$: $x = 9 - 12 = -3$, $y = -27 + 9 = -18$

N2009 P2 Q1

N2009 P2 Q1

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