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Conics and Foci Solutions

• Focal points, or foci, $F_1$ and $F_2$, are two fixed points on the major axis of an ellipse or hyperbola. 

  • For an ellipse, the sum of the distances from any point, P, on an ellipse to the foci is a constant value: $d_1+d_2=\mathrm{constant}$

  • For a hyperbola, the absolute value of the difference of the distance from any point, P, on a hyperbola to the foci is a constant value: $\left|d_1–d_2\right|=\mathrm{constant}$

• Focal length, c, is the distance from the center $(h, k)$ of an ellipse or hyperbola to one focal point.

• The midpoint formula can be used to locate the center if the foci are given.

• When $a>0$ and $b>0$:

  • The major axis (ellipse) or the transverse axis (hyperbola) is a segment with length 2a. 

  • The minor axis (ellipse) or the conjugate axis (hyperbola) is a segment with length 2b.

• You must first determine if the major or the transverse axis is vertical or horizontal to work with focal points.

	Horizontal Ellipse	Horizontal Hyperbola
Horizontal major axis with center $\mathbf{(}\mathit{h}\mathbf{,}\mathbf{ }\mathit{k}\mathbf{)}$	$\frac{{\left(x–h\right)}^2}{a^2}+\frac{{\left(y–k\right)}^2}{b^2}=1$	$\frac{{\left(x–h\right)}^2}{a^2}–\frac{{\left(y–k\right)}^2}{b^2}=1$
Horizontal foci	Have the same y-coordinate: $\left(h\pm c, k\right)$	Have the same y-coordinate: $\left(h\pm c, k\right)$
Axes	Major axis: 2a Minor axis: 2b $a>b,$ always	Transverse axis: 2a Conjugate axis: 2b The transverse axis is represented by the term with x in the numerator.
Focal length	${\left(\mathrm{major}\right)}^2–{\left(\mathrm{minor}\right)}^2={\left(\mathrm{focal}\right)}^2$ or $a^2–b^2=c^2$	${\left(\mathrm{transverse}\right)}^2+{\left(\mathrm{conjugate}\right)}^2={\left(\mathrm{focal}\right)}^2$ or $a^2+b^2=c^2$

	Vertical Ellipse	Vertical Hyperbola
Major vertical axis with center $\mathbf{(}\mathit{h}\mathbf{,}\mathbf{ }\mathit{k}\mathbf{)}$	$\frac{{\left(x–h\right)}^2}{b^2}+\frac{{\left(y–k\right)}^2}{a^2}=1$	$\frac{{\left(x–h\right)}^2}{a^2}–\frac{{\left(y–k\right)}^2}{b^2}=1$
Vertical foci	Have the same x-coordinate: $\left(h, k\pm c\right)$	Have the same x-coordinate: $\left(h, k\pm c\right)$
Axes	Major axis: 2a Minor axis: 2b $a>b,$ always	Transverse axis: 2a Conjugate axis: 2b The transverse axis is represented by the term with y in the numerator.
Focal length	${\left(\mathrm{major}\right)}^2–{\left(\mathrm{minor}\right)}^2={\left(\mathrm{focal}\right)}^2$ or $a^2–b^2=c^2$	${\left(\mathrm{transverse}\right)}^2+{\left(\mathrm{conjugate}\right)}^2={\left(\mathrm{focal}\right)}^2$ or $a^2+b^2=c^2$

Example 1

State whether the equation is a vertical or horizontal ellipse or hyperbola. Then determine the foci.

Example 2

Determine the equation of the ellipse in standard form when the length of its vertical minor axis is 14 units and the focal points are at $\mathbf{(}\mathbf{–}\mathbf{30}\mathbf{,}\mathbf{ }\mathbf{–}\mathbf{8}\mathbf{)}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{(}\mathbf{18}\mathbf{,}\mathbf{ }\mathbf{–}\mathbf{8}\mathbf{)}\mathbf{.}$

Equation: $\frac{{\left(x+6\right)}^2}{625}+\frac{{\left(y+8\right)}^2}{49}=1$

Example 3

Determine the equation of the hyperbola in standard form with vertices at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{5}\mathbf{)}$ and $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{–}\mathbf{1}\mathbf{)}$ and focal points at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{7}\mathbf{)}$ and $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{–}\mathbf{3}\mathbf{)}.$

Equation: $\frac{{\left(y–2\right)}^2}{9}–\frac{{\left(x–1\right)}^2}{16}=1$