WAEC 2018 · Paper 2 · Q10

(a)
Find the equation of the normal to the curve $x^2 + xy + 2y^2 = 8$ at the point $(-3, 1)$ .
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$5y + x - 2 = 0$
(b)
Find the equation of a line passing through the midpoint of the line joining $A(5, 1)$ and $B(-1, -5)$ and perpendicular to line $AB$ .
Show the answer
$y + x = 0$

Try it on a graph

The curve x² + xy + 2y² = 8 (upper and lower halves) and the normal at (−3, 1).

Worked solution (try it first)

(a)

Differentiate implicitly: $2x + \left(y + x\dfrac{dy}{dx}\right) + 4y\dfrac{dy}{dx} = 0$ .
Collect: $(x + 4y)\dfrac{dy}{dx} = -(2x + y)$ , so $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 4y}$ .
At $(-3, 1)$ : $\dfrac{dy}{dx} = -\dfrac{-6 + 1}{-3 + 4}$
$= 5$ , so the normal's gradient is $-\frac15$ .
The normal: $y - 1 = -\frac15(x + 3)$ .
Multiply by 5: $5y - 5 = -x - 3$ , so $5y + x - 2 = 0$ .

(b)

The midpoint of $AB$ is $\left(\dfrac{5 - 1}{2}, \dfrac{1 - 5}{2}\right) = (2, -2)$ .
The gradient of $AB$ is $\dfrac{-5 - 1}{-1 - 5} = 1$ , so a perpendicular line has gradient $-1$ .
The line: $y + 2 = -(x - 2)$ , so $y = -x$ , that is $y + x = 0$ .

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