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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Parametric equations แพ็กฝึกเฉพาะหัวข้อ · สมการพาราเมตริก

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Cartesian form, tangents & normals, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

A curve has parametric equations $x=2t+1$, $y=t^2$. Find a cartesian equation of the curve.

Mark schemeshow ▾
M1 $t=\dfrac{x-1}{2}$  M1 substitute into $y=t^2$  A1 $y=\dfrac{(x-1)^2}{4}$.
Q2. 4 marks

A curve has parametric equations $x=2\cos\theta$, $y=5\sin\theta$, $0\le\theta<2\pi$. Use a trigonometric identity to find a cartesian equation of the curve.

Mark schemeshow ▾
B1 $\cos\theta=\dfrac{x}{2}$  B1 $\sin\theta=\dfrac{y}{5}$  M1 $\cos^2\theta+\sin^2\theta=1$  A1 $\dfrac{x^2}{4}+\dfrac{y^2}{25}=1$ (an ellipse).
Q3. 4 marks

A curve has parametric equations $x=2t-4$, $y=t^2-1$. Find the coordinates of the points where the curve crosses (a) the $x$-axis, (b) the $y$-axis.

Mark schemeshow ▾
(a) M1 $y=0\Rightarrow t^2-1=0\Rightarrow t=\pm1$  A1 $t=1$: $(-2,0)$; $t=-1$: $(-6,0)$.
(b) M1 $x=0\Rightarrow 2t-4=0\Rightarrow t=2$  A1 $y=4-1=3$, point $(0,3)$.
Q4. 5 marks

A curve has parametric equations $x=t^2$, $y=t^3-4t$. Find the value of $\dfrac{dy}{dx}$ at the point where $t=2$.

Mark schemeshow ▾
B1 $\dfrac{dx}{dt}=2t$  B1 $\dfrac{dy}{dt}=3t^2-4$  M1 $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2-4}{2t}$  M1 substitute $t=2$: $\dfrac{12-4}{4}$  A1 $\dfrac{dy}{dx}=2$.
Q5. 5 marks

A curve has parametric equations $x=2\cos\theta$, $y=3\cos2\theta$, $0\le\theta\le\pi$. Find a cartesian equation of the curve, giving the range of values of $x$ for which it is defined.

Mark schemeshow ▾
B1 $\cos\theta=\dfrac{x}{2}$  M1 $\cos2\theta=2\cos^2\theta-1$  M1 $y=3\left(2\cdot\dfrac{x^2}{4}-1\right)$  A1 $y=\dfrac{3x^2}{2}-3$  B1 $-2\le x\le 2$ (since $-1\le\cos\theta\le1$).
Q6. 6 marks

A curve has parametric equations $x=t^2$, $y=2t$. Find an equation of the tangent to the curve at the point where $t=3$, giving your answer in the form $ax+by+c=0$ where $a$, $b$, $c$ are integers.

Mark schemeshow ▾
B1 point: $(9,6)$  M1 $\dfrac{dx}{dt}=2t$, $\dfrac{dy}{dt}=2$  A1 $\dfrac{dy}{dx}=\dfrac{2}{2t}=\dfrac1t=\dfrac13$ at $t=3$  M1 $y-6=\tfrac13(x-9)$  A1 correct unsimplified line  A1 $x-3y+9=0$.
Q7. 6 marks

A curve has parametric equations $x=t^3-3t$, $y=t^2$. Find the coordinates of (a) the point where the tangent to the curve is parallel to the $x$-axis, (b) the points where the tangent is parallel to the $y$-axis.

Mark schemeshow ▾
B1 $\dfrac{dx}{dt}=3t^2-3$, $\dfrac{dy}{dt}=2t$.
(a) M1 parallel to $x$-axis: $\dfrac{dy}{dt}=0$ (with $\dfrac{dx}{dt}\neq0$): $2t=0\Rightarrow t=0$  A1 point $(0,0)$.
(b) M1 parallel to $y$-axis: $\dfrac{dx}{dt}=0$: $3t^2-3=0$  A1 $t=\pm1$  A1 $t=1$: $(-2,1)$; $t=-1$: $(2,1)$.
Q8. 7 marks

The curve $C$ has parametric equations $x=4\cos\theta$, $y=2\sin\theta$, $0\le\theta<2\pi$. Find an equation of the normal to $C$ at the point where $\theta=\dfrac{\pi}{4}$.

Mark schemeshow ▾
B1 point: $\left(2\sqrt2,\ \sqrt2\right)$  M1 $\dfrac{dx}{d\theta}=-4\sin\theta$, $\dfrac{dy}{d\theta}=2\cos\theta$  A1 $\dfrac{dy}{dx}=\dfrac{2\cos\theta}{-4\sin\theta}=-\dfrac{\cos\theta}{2\sin\theta}$  A1 at $\theta=\tfrac{\pi}{4}$: $\dfrac{dy}{dx}=-\tfrac12$  M1 normal gradient $=-1\div\left(-\tfrac12\right)=2$  M1 $y-\sqrt2=2\left(x-2\sqrt2\right)$  A1 $y=2x-3\sqrt2$.
Q9. 7 marks

The curve $C$ has parametric equations $x=1+2t$, $y=\dfrac{4}{t}$, $t\neq0$. (a) Find a cartesian equation of $C$. (b) Find an equation of the tangent to $C$ at the point where $t=2$, giving your answer in the form $ax+by+c=0$ where $a$, $b$, $c$ are integers.

Mark schemeshow ▾
(a) M1 $t=\dfrac{x-1}{2}$, substitute into $y=\dfrac4t$  A1 $y=\dfrac{8}{x-1}$, $x\neq1$.
(b) B1 point: $(5,2)$  M1 $\dfrac{dx}{dt}=2$, $\dfrac{dy}{dt}=-\dfrac{4}{t^2}$  A1 $\dfrac{dy}{dx}=-\dfrac{2}{t^2}=-\tfrac12$ at $t=2$  M1 $y-2=-\tfrac12(x-5)$  A1 $x+2y-9=0$.
Q10. 8 marks

The curve $C$ has parametric equations $x=2\sin\theta$, $y=\cos2\theta$, $-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$. (a) Show that a cartesian equation of $C$ is $y=1-\dfrac{x^2}{2}$, stating the range of values of $x$. (b) Find the value of $\dfrac{dy}{dx}$ at the point where $\theta=\dfrac{\pi}{6}$. (c) Find an equation of the normal to $C$ at this point, and the coordinates of the point where this normal crosses the $y$-axis.

Mark schemeshow ▾
(a) M1 $\cos2\theta=1-2\sin^2\theta$  M1 $\sin\theta=\dfrac{x}{2}$, so $y=1-2\cdot\dfrac{x^2}{4}$  A1 $y=1-\dfrac{x^2}{2}$ ∎  B1 $-2\le x\le 2$.
(b) M1 $\dfrac{dy}{dx}=\dfrac{-2\sin2\theta}{2\cos\theta}$; at $\theta=\tfrac{\pi}{6}$: $\dfrac{-2\sin\frac{\pi}{3}}{2\cos\frac{\pi}{6}}=\dfrac{-\sqrt3}{\sqrt3}$  A1 $\dfrac{dy}{dx}=-1$ (point is $\left(1,\tfrac12\right)$).
(c) M1 normal gradient $=1$: $y-\tfrac12=1\cdot(x-1)$  A1 $y=x-\tfrac12$; crosses $y$-axis at $\left(0,-\tfrac12\right)$.