Normal to a parabola

Conic sections

A-level Further Maths (9FM0) · Further Pure 1

ORIGINAL

Gradient of the curve, perpendicular gradient, then solve with the parabola — one root is the point you started from.

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Question

The parabola

C

has equation

y^2 = 12x

. The point

P(12, 12)

lies on

C

. The normal to

C

P

meets

C

again at the point

Q

. Find the coordinates of

Q

Every step worked, with the reasoning.

1
$2y\dfrac{dy}{dx} = 12 \Rightarrow \dfrac{dy}{dx} = \dfrac{6}{y}$
Differentiate $y^2 = 12x$ implicitly.
2
At $P$ : $\dfrac{dy}{dx} = \dfrac{6}{12} = \dfrac12$ , so the normal has gradient $-2$
The normal is perpendicular to the tangent: $m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}}$ .
3
$y - 12 = -2(x - 12) \Rightarrow y = 36 - 2x$
Point-gradient form through $P(12, 12)$ .
4
$x = \dfrac{36 - y}{2}$ , so $y^2 = 6(36 - y) \Rightarrow y^2 + 6y - 216 = 0$
Substitute the normal into $y^2 = 12x$ ; eliminating $x$ keeps the algebra linear in $x$ .
5
$(y - 12)(y + 18) = 0 \Rightarrow y = 12$ or $y = -18$
$y = 12$ is $P$ itself — it must be a root, which is a useful check on the quadratic.
6
$y = -18 \Rightarrow x = \dfrac{36 + 18}{2} = 27$
Back-substitute; check: $(-18)^2 = 324 = 12 \times 27$ .

Answer: $Q(27,\ -18)$