June Revision Exercise 3 Q1

By KS Teng

(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 3JC Mathematics
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