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Ellipses and Hyperbolas

In this lesson you’ll learn the standard equations for ellipses and hyperbolas and how to identify their key features.

An ellipse is an oval-shaped curve. The standard equation with center (h, k) is:

Horizontal major axis (wider than tall, a > b):

(xh)2a2+(yk)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1

Vertical major axis (taller than wide, a > b):

(xh)2b2+(yk)2a2=1\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1

Key features: center (h, k), vertices along the major axis, co-vertices along the minor axis.

A hyperbola has two separate branches. The standard equations are:

Horizontal transverse axis (branches open left/right):

(xh)2a2(yk)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

Vertical transverse axis (branches open up/down):

(yk)2a2(xh)2b2=1\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1

Key features: center (h, k), vertices along the transverse axis, asymptotes that the branches approach.

The left graph shows the green ellipse centered at (2, −1) with a = 3 (horizontal) and b = 2 (vertical). The gold dot is the center, and the blue dots are the vertices at (−1, −1) and (5, −1). The right graph shows the purple hyperbola centered at (−3, 5) with two branches opening left and right. The dashed lines are the asymptotes that the branches approach but never touch.

1. Identify the conic and center

(x2)29+(y+1)24=1\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1

Both terms positive → ellipse. Center: (2, −1). Since 9 > 4, the larger denominator is under x², so a = 3 (horizontal major axis) and b = 2.

2. Identify the conic

(x+3)216(y5)29=1\frac{(x + 3)^2}{16} - \frac{(y - 5)^2}{9} = 1

Opposite signs → hyperbola. The positive term is under x², so the transverse axis is horizontal. Center: (−3, 5). a = 4, b = 3.

  • Ellipses: planetary orbits (Kepler’s laws), whispering galleries, certain sports stadiums, and gears.
  • Hyperbolas: navigation systems (LORAN), cooling towers at power plants, some telescope designs, and hyperbolic mirrors.

Example: the orbits of planets around the sun are ellipses, with the sun at one focus.

The equation $\frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$ is a:
The equation $\frac{(x+3)^2}{25} - \frac{(y-1)^2}{16} = 1$ is a:
In the ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, if $a > b$, the major axis is:
The center of $\frac{(x+4)^2}{16} - \frac{(y-3)^2}{9} = 1$ is:
For the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$, what is the length of the semi-major axis?