Edexcel A-Level Mathematics A2 P4.3.1 Parametric Equations of Curves and Conversion Between QuestionsPractise Edexcel IAL P4.3.1 parametric equations: eliminate parameters, use trigonometric identities, and find coordinates or intersections.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsEliminate the parameter to form a Cartesian equation with valid restrictionsFind curve coordinates or intersections by substituting parameter values and line equationsUse trigonometric identities to convert sine and cosine parameters without trig terms
Edexcel A-Level Mathematics A2 P4.3.1 Parametric Equations of Curves and Conversion Between Questions question 1[Maximum number: 5]Figure 4Figure 4 shows a sketch of the curve C with parametric equationsx=secty=3tan(t+π3)π6<t<π2x=\sec t \quad y=\sqrt{3} \tan \left(t+\frac{\pi}{3}\right) \quad \frac{\pi}{6}<t<\frac{\pi}{2}x=secty=3tan(t+3π)6π<t<2πShow that all points on C satisfy the equationy=Ax2+B3x2−34−3x2y=\frac{A x^{2}+B \sqrt{3 x^{2}-3}}{4-3 x^{2}}y=4−3x2Ax2+B3x2−3where A and B are constants to be found.Mark as masteredShow Answery=3tant±tanπ31±tanttanπ3y=\sqrt{3} \frac{\tan t \pm \tan \frac{\pi}{3}}{1 \pm \tan t \tan \frac{\pi}{3}}y=31±tanttan3πtant±tan3πx2=sec2t=1+tan2t⇒tant=x2−1⇒y=3x2−1+31−3x2−1x^{2}=\sec ^{2} t=1+\tan ^{2} t \Rightarrow \tan t=\sqrt{x^{2}-1} \Rightarrow y=\sqrt{3} \frac{\sqrt{x^{2}-1}+\sqrt{3}}{1-\sqrt{3} \sqrt{x^{2}-1}}x2=sec2t=1+tan2t⇒tant=x2−1⇒y=31−3x2−1x2−1+3=3x2−1+31−3x2−1×1+3x2−11+3x2−1=.=\sqrt{3} \frac{\sqrt{x^{2}-1}+\sqrt{3}}{1-\sqrt{3} \sqrt{x^{2}-1}} \times \frac{1+\sqrt{3} \sqrt{x^{2}-1}}{1+\sqrt{3} \sqrt{x^{2}-1}}=.=31−3x2−1x2−1+3×1+3x2−11+3x2−1=..=3x2−1+3(x2−1)+3+3x2−11−(3x2−3)=..x2−1+..x24−3x2=\sqrt{3} \frac{\sqrt{x^{2}-1}+\sqrt{3}\left(x^{2}-1\right)+\sqrt{3}+3 \sqrt{x^{2}-1}}{1-\left(3 x^{2}-3\right)}=\frac{. . \sqrt{x^{2}-1}+. . x^{2}}{4-3 x^{2}}=31−(3x2−3)x2−1+3(x2−1)+3+3x2−1=4−3x2..x2−1+..x2=3x2+43x2−34−3x2=\frac{3 x^{2}+4 \sqrt{3 x^{2}-3}}{4-3 x^{2}}=4−3x23x2+43x2−3(5)Add to Test