(a) Find the equation of the tangent to curve \(\frac{x^{2}}{4} + y^{2} = 1\) at point \(1, \frac{\sqrt{3}}{2}\). (b) Express \(\frac{3x + 2}{x^{2} + x - 2}...
Question 1 Report
(a) Find the equation of the tangent to curve \(\frac{x^{2}}{4} + y^{2} = 1\) at point \(1, \frac{\sqrt{3}}{2}\).
(b) Express \(\frac{3x + 2}{x^{2} + x - 2}\) in partial fractions.
Answer Details
(a) Differentiate \(\dfrac{x^2}{4}+y^2=1\) implicitly:
\[\frac{2x}{4}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{4y}\]
At \(\left(1,\tfrac{\sqrt3}{2}\right)\): \(\dfrac{dy}{dx}=-\dfrac{1}{4\cdot\frac{\sqrt3}{2}}=-\dfrac{1}{2\sqrt3}=-\dfrac{\sqrt3}{6}\).
Tangent: \(y-\dfrac{\sqrt3}{2}=-\dfrac{\sqrt3}{6}(x-1)\). Multiplying through by 6 and simplifying:
\[\sqrt3\,x+6y=4\sqrt3\quad\text{or}\quad x+2\sqrt3\,y=4\]
(b) Factorise the denominator: \(x^2+x-2=(x+2)(x-1)\). Write
\[\frac{3x+2}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\Rightarrow 3x+2=A(x-1)+B(x+2)\]
Put \(x=1\): \(5=3B\Rightarrow B=\tfrac53\). Put \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).
\[\frac{3x+2}{x^2+x-2}=\frac{4}{3(x+2)}+\frac{5}{3(x-1)}\]
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