WAEC 2013 · Paper 2 · Q9✱✱

Find the:

(a)
gradients of the tangents to the curve $x^2 + y^2 - 2x - 6y + 5 = 0$ at the points where $x = 0$ ;
Separate values with commas, e.g. 3, −2
(b)
equations of the tangents to the curve in (a).
Show the answer
$x + 2y - 2 = 0$ and $x - 2y + 10 = 0$

Worked solution (try it first)

(a)

Put $x = 0$ in the equation: $y^2 - 6y + 5 = 0$ .
Factorise: $(y - 1)(y - 5) = 0$ , so the points are $(0, 1)$ and $(0, 5)$ .
Differentiate implicitly: $2x + 2y\dfrac{dy}{dx} - 2 - 6\dfrac{dy}{dx} = 0$ .
Collect the $\dfrac{dy}{dx}$ terms: $(2y - 6)\dfrac{dy}{dx} = 2 - 2x$ , so $\dfrac{dy}{dx} = \dfrac{1 - x}{y - 3}$ .
At $(0, 1)$ the gradient is $\dfrac{1}{-2} = -\dfrac12$ .
At $(0, 5)$ the gradient is $\dfrac{1}{2}$ .

(b)

At $(0, 1)$ : $y - 1 = -\frac12(x - 0)$ .
Multiply by 2 and rearrange to get $x + 2y - 2 = 0$ .
At $(0, 5)$ : $y - 5 = \frac12(x - 0)$ .
Multiply by 2 and rearrange to get $x - 2y + 10 = 0$ .
The tangents are $x + 2y - 2 = 0$ and $x - 2y + 10 = 0$ .

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