Stepwise
Start practisingPractise

Conic sections · Lesson 1 of 1

The parabola and the ellipse

The parabola y² = 4ax with its focus and directrix, the ellipse x²/a² + y²/b² = 1, and tangents to them by implicit differentiation.

14 minYou should already know: Coordinate geometry & circles
  1. 1

A circle is the set of points at a fixed distance from its centre (see circles). Two more curves come from similar rules.

The parabola

A parabola is the set of points that are the same distance from a fixed point, the focus, and a fixed line, the directrix. With the focus at (a,0)(a, 0) and the directrix x=−ax = -a, the equation is

F(a, 0)Px = −a
The parabola y² = 4axfocus (a, 0); directrix x = −a; PF = PN
y2=4axy^2 = 4ax

To find the focus and directrix, compare the equation with y2=4axy^2 = 4ax to find aa.

Focus and directrixSet a, then slide P along the curve
−5−3−113579−6−4−2246xyFP
4.167PF (to the focus)4.167PN (to the directrix)
y² = 4(1.5)x = 6x: focus (1.5, 0), directrix x = −1.5. P is (2.67, 4). PF = √((2.67 − 1.5)² + 4²) = 4.167 and PN = 2.67 + 1.5 = 4.167: always equal.

The ellipse

An ellipse with centre at the origin, crossing the xx-axis at ±a\pm a and the yy-axis at ±b\pm b, has equation

ab(a, 0)(0, b)
The ellipsex²/a² + y²/b² = 1
x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Tangents

To find the gradient of either curve at a point, differentiate implicitly (see implicit differentiation): y2y^2 differentiates to 2ydydx2y\dfrac{dy}{dx}.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q12 (a)

Find the equation of the tangent to the curve x24+y2=1\dfrac{x^2}{4} + y^2 = 1 at the point (1,32)\left(1, \dfrac{\sqrt3}{2}\right).

  1. Differentiate

    • x2+2ydydx=0{\frac{x}{2} + 2y\frac{dy}{dx} = 0}.
    • So dydx=−x4y{\frac{dy}{dx} = -\frac{x}{4y}}.

    Think first. Differentiate each term with respect to x.

  2. The gradient at the point

    • dydx=−14×32=−123{\frac{dy}{dx} = -\frac{1}{4 \times \frac{\sqrt3}{2}} = -\frac{1}{2\sqrt3}}.

  3. The tangent

    • y−32=−123(x−1){y - \frac{\sqrt3}{2} = -\frac{1}{2\sqrt3}(x - 1)}.
    • Multiply by 23{2\sqrt3}: 23y−3=−x+1{2\sqrt3y - 3 = -x + 1}.
    • So x+23y−4=0{x + 2\sqrt3y - 4 = 0}.

Your turn

NECO 2023 · Paper 1 · Q28

Find the focus and directrix of the parabola y2=64xy^2 = 64x.

Worked solution (try it first)
  1. Compare y2=64xy^2 = 64x with the standard form y2=4axy^2 = 4ax: 4a=644a = 64, so a=16a = 16.
  2. The focus is (a,0)=(16,0)(a, 0) = (16, 0) and the directrix is the line x=−ax = -a, that is x=−16x = -16, option C.

Report a problem with this question