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Mathematics · Ch 5 — Two Dimensional Analytical Geometry-II

Parabola

5.3.2

Parabola

Since e=1e=1 for a parabola, a parabola is simply the set of points equidistant from a fixed focus and a fixed directrix.

(i) Standard form, vertex at the origin. Let SS be the focus and ℓ\ell the directrix; draw SZ⊥ℓSZ\perp\ell, and take the line SZSZ (produced) as the xx-axis with the perpendicular bisector of SZSZ as the yy-axis, so the origin OO is their intersection. Let SZ=2aSZ=2a, so S=(a,0)S=(a,0) and ℓ:x+a=0\ell:x+a=0. For a moving point P(x,y)P(x,y) with PM⊥ℓPM\perp\ell, the defining condition SP=PMSP=PM (since e=1e=1) gives

(x−a)2+y2=(x+a)2  ⟹  y2=4ax,(x-a)^2+y^2=(x+a)^2 \implies y^2=4ax,

the parabola in standard form. The three other origin-vertex orientations follow by symmetry: y2=−4axy^2=-4ax (opens left), x2=4ayx^2=4ay (opens up), x2=−4ayx^2=-4ay (opens down).

Vocabulary (Definition 5.3). The line through the focus perpendicular to the directrix is the axis; where the axis meets the curve is the vertex; any chord through the focus is a focal chord; the focal chord perpendicular to the axis is the latus rectum, and its endpoints for y2=4axy^2=4ax are (a,±2a)(a,\pm2a) (found by substituting x=ax=a), so its length is 4a4a. The parabola y2=4axy^2=4ax is symmetric about the xx-axis (replacing yy by −y-y leaves the equation unchanged) and lies entirely on the side x≥0x\ge0.

(ii) Vertex at (h,k)(h,k). Shifting the origin to (h,k)(h,k): when the axis is parallel to the xx-axis, the equation is (y−k)2=±4a(x−h)(y-k)^2=\pm4a(x-h); when parallel to the yy-axis, (x−h)2=±4a(y−k)(x-h)^2=\pm4a(y-k). Summarised (with a>0a>0 throughout):

EquationVertexFocusAxisDirectrixLatus rectum
(y−k)2=4a(x−h)(y-k)^2=4a(x-h)(h,k)(h,k)(h+a,k)(h+a,k)y=ky=kx=h−ax=h-a4a4a
Figure 5.17–5.18Parts of a parabola

What this figure shows. Focus SS, directrix ℓ\ell, vertex, axis and the latus rectum LL′LL' through the focus perpendicular to the axis, for y2=4axy^2=4ax. …

Figure 5.19–5.22The four (h,k)-vertex parabola orientations

What this figure shows. Four small sketches — opening right, left, up and down — each showing the vertex (h,k)(h,k), the shifted focus and directrix for (y−k)2=±4a(x−h)(y-k)^2=\pm4a(x-h) and (x−h)2=±4a(y−k)(x-h)^2=\pm4a(y-k). …