2019 EJC Promo Q3

2019 EJC Promo Q3

6 marks
Promo

The curve \(C\) has equation \({{a}^{2}}{{x}^{2}}-{{y}^{2}}-2\left( ax-2y+2 \right)=0\) where \(a\) is a constant such \(a\in \mathbb{R}\backslash \left\{ 0 \right\}\).

Show that \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{a\left( ax-1 \right)}{y-2}\).

Hence, find the coordinates of the points on \(C\) at which the tangent is parallel to \(y\)-axis.[6]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{a(ax-1)}{y-2}\). The tangents parallel to the \(y\)-axis occur at \((0,2)\) and \(\left(\dfrac2a,2\right)\).

Need help? Join our JC Math tuition classes.

Learn more