Skill: conics-ellipses-hyperbolas

Ellipses & Hyperbolas

Graph ellipses and hyperbolas with confidence: find centers, vertices, foci with the right c² relationship, and write asymptote equations.

1 Ellipses: two foci, one constant sum

An ellipse is the set of points whose distances to two foci add to a constant (2a).

Ellipse: standard form

(x − h)²a² + (y − k)²b² = 1 · c² = a² − b²

Center (h, k). The larger denominator is a² and sets the major (transverse) axis: if a² sits under (x − h)², vertices are (h ± a, k) and foci (h ± c, k) with c² = a² − b² (minus for ellipses).

Left: ellipse x²/25 + y²/9 = 1 — a = 5, b = 3, c = √(25−9) = 4. Right: hyperbola x²/9 − y²/16 = 1 — a = 3, b = 4, c = √(9+16) = 5, asymptotes y = ±(4/3)x.

2 Hyperbolas: constant difference, plus asymptotes

A hyperbola is the set of points whose distances to two foci have a constant difference (2a). Two branches, plus asymptote guides.

Hyperbola: standard form

(x − h)²a² − (y − k)²b² = 1 · c² = a² + b²

The positive term’s denominator is a² and sets the transverse axis: here vertices (h ± a, k), foci (h ± c, k) with c² = a² + b² (plus for hyperbolas). Asymptotes: y − k = ±(b/a)(x − h). If the y-term is positive instead, everything runs vertically.

3 Worked examples

Identify, then locate — each claim verified independently.

Example 1 Ellipse or hyperbola?

Classify x²/25 + y²/9 = 1.

Plus sign between squares → ellipse.

Example 2 Ellipse or hyperbola?

Classify x²/9 − y²/16 = 1.

Minus sign between squares → hyperbola.

Example 3 Focal distance: ellipse

For x²/25 + y²/9 = 1, find c.

c² = a² − b² = 25 − 9 = 16, so c = 4.

Example 4 Focal distance: hyperbola

For x²/9 − y²/16 = 1, find c.

c² = a² + b² = 9 + 16 = 25, so c = 5.

Example 5 Foci of an ellipse

Find the foci of x²/25 + y²/9 = 1.

c = 4, major axis horizontal: foci (−4, 0) and (4, 0).

Example 6 Foci of a shifted hyperbola

Find the foci of (x − 1)²/9 − (y + 2)²/16 = 1.

Center (1, −2), c = 5, transverse axis horizontal: foci (−4, −2) and (6, −2).

Example 7 Asymptotes

Find the asymptotes of x²/9 − y²/16 = 1.

y = ±(b/a)x = ±(4/3)x: y = (4/3)x and y = −(4/3)x.

4 Common mistakes

The two errors that cost the most points.

Using a² + b² = c² for hyperbolas

Wrong: c² = a² − b² on a hyperbola, giving c = √(9−16).
Right: hyperbolas use plus: c² = a² + b² = 9 + 16 = 25, c = 5. Minus is for ellipses.

Swapping transverse and conjugate axes

Wrong: on (y−1)²/16 − (x+2)²/9 = 1, putting vertices at (−2±3, 1).
Right: the positive term’s denominator is a² = 16, so a = 4 and the transverse axis is vertical: vertices (−2, 1±4).

5 Key vocabulary

Words to know

Transverse axis — the axis through the vertices; its half-length is a.
Conjugate axis — the perpendicular axis through the center; its half-length is b.
Focal distance (c) — center-to-focus distance: c² = a² − b² (ellipse), a² + b² (hyperbola).
Asymptotes — the guide lines a hyperbola approaches: y − k = ±(b/a)(x − h).

6 Quick check

Try these before moving on — click to reveal each answer.

For x²/49 + y²/24 = 1, find c.

c² = 49 − 24 = 25, so c = 5.

For y²/16 − x²/9 = 1, find c and the transverse axis direction.

c² = 16 + 9 = 25, c = 5; the y-term is positive so the transverse axis is vertical.

Give the asymptotes of (x − 2)²/25 − (y + 1)²/144 = 1.

b/a = 12/5: y + 1 = ±(12/5)(x − 2).

7 Next steps

Now drill the skill with endless randomized problems.

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