June Revision Exercise 3 Q1

(i)

$2x^2 + 8xy + 5y^2 = -3$

$4x + (8x \frac{dy}{dx} + 8y) + 10y \frac{dy}{dx}=0$

$\frac{dy}{dx} = \frac{2x+4y}{4x+5y}$

For tangents to be parallel to the $x$ -axis, $\frac{dy}{dx}=0$

$\Rightarrow x=-2y$

Sub $x=-2y$ into $2x^2 + 8xy + 5y^2 = -3$

$y= \pm 1$

Thus, equations of tangents which are parallel to $x$ -axis are $y= 1 \text{ or } y=-1$

(ii)

$\frac{dy}{dx}=\frac{dy}{dx}\bullet \frac{dx}{dt} = \frac{2[3+2(-1.15)]}{4(3)+5(-1.15)} = -0.488$ units per second.

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June Revision Exercise 3 Q1

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