Hyperbola

A hyperbola consists of two curves opening in opposite directions. Just like one of its conic partners, the ellipse, a hyperbola also has two foci and is defined as the set of points where the absolute value of the difference of the distances to the two foci is constant. Thus, every point \(X\) of a hyperbola whose foci are points \(F\) and \(F'\) would satisfy \(\lvert\overline{XF}-\overline{XF'}\rvert=k,\) where \(k\) is a constant.

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A hyperbola consists of a center, an axis, two vertices, two foci, and two asymptotes. A hyperbola's axis is the line that passes through the two foci, and the center is the midpoint of the two foci. The two vertices are where the hyperbola meets with its axis. On the coordinate plane, we most often use the \(x\)- or \(y\)-axis as the hyperbola's axis. The equation for the hyperbola in each of these cases is as follows:

The equation of a hyperbola whose axis is the \(x\)-axis and whose center is the origin \(O\) is

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.\]

Every point \(X\) of the hyperbola satisfies \(\lvert\overline{XF}-\overline{XF'}\rvert=2a,\) where the coordinates of the two foci are \(F(c,0)\) and \(F'(-c,0),\) and \(c^2=a^2+b^2.\) The coordinates of the vertices are \((a,0)\) and \((-a,0).\)

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The equation of a hyperbola whose axis is the \(y\)-axis and whose center is the origin \(O\) is

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1.\]

Every point \(X\) of the hyperbola satisfies \(\lvert\overline{XF}-\overline{XF'}\rvert=2b,\) where the coordinates of the two foci are \(F(0,c)\) and \(F'(0,-c),\) and \(c^2=a^2+b^2.\) The coordinates of the vertices are \((0,b)\) and \((0,-b).\)

By translating these equations, we can express any hyperbola on the coordinate plane whose axis is parallel to either the \(x\) or \(y\)-axis.

Now let's discuss the asymptotes of a hyperbola. Every hyperbola has two asymptotes that are symmetrical about the hyperbola's axis. For a hyperbola whose equation is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm1,\) the equations of the asymptotes are

\[y=\pm\frac{b}{a}x.\]

Imagine taking the limit of \(x\rightarrow\infty.\) Then \(y\) will also increase indefinitely, and the 1 on the right-hand side will be eliminated. Hence the ratio between \(y\) and \(x\) will either become \(b:a\) or \(-b:a.\)

Find the two foci of the hyperbola

\[\frac{x^2}{3}-\frac{y^2}{7}=1.\]

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The given hyperbola has the origin \(O=(0,0)\) as its center and the \(x\)-axis as its axis. Then the two foci are given by \(F=(c,0)\) and \(F'=(-c,0),\) where \(3+7=c^2.\) Hence, \(c=\sqrt{10}\) and the two foci are \(F=\big(\sqrt{10},0\big)\) and \(F'=\big(-\sqrt{10},0\big).\ _\square\)

What is the equation of a hyperbola whose foci are \(F=\big(\sqrt{5},0\big)\) and \(F'=\big(-\sqrt{5},0\big),\) and the absolute value of the difference of the distances to the two foci is 4?

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Since the two foci are given as \(F=\big(\sqrt{5},0\big)\) and \(F'=\big(-\sqrt{5},0\big),\) the center of the hyperbola is the origin \(O=(0,0)\) and its axis is the \(x\)-axis. Hence we have

\[\begin{align} 2a&=4\\ \Rightarrow a&=2\\ a^2+b^2&=5\\ \Rightarrow b^2&=1\\ \Rightarrow\frac{x^2}{4}-y^2&=1.\ _\square \end{align}\]

What is the equation of a hyperbola whose foci are \(F=(0,5)\) and \(F'=(0,-5),\) and the absolute value of the difference of the distances to the two foci is 6?

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Since the two foci are given as \(F=(0,5)\) and \(F'=(0,-5),\) the center of the hyperbola is the origin \(O=(0,0)\) and its axis is the \(y\)-axis. Hence we have

\[\begin{align} 2b&=6\\ \Rightarrow b&=3\\ a^2+b^2&=5^2\\ \Rightarrow a^2&=16\\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}&=-1.\ _\square \end{align}\]

What is the equation of a hyperbola whose foci are \(F=(7,3)\) and \(F'=(1,3),\) and the absolute value of the difference of the distances to the two foci is 4?

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Since the axis is the line that passes through the two foci, the axis of the given hyperbola is \(y=3.\) The center of the hyperbola is the midpoint of the two foci, which is \((4,3)\) in this case. We can think of the hyperbola as a parallel translation of four units to the right and 3 units upward from \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.\) Since the absolute value of the difference of the distances to the two foci is 4, it must be true that \(2a=4,\) or \(a=2.\) Since the distance between the two foci is 6, we know that \(c=3.\) Hence we have

\[\begin{align} a^2+b^2&=c^2\\ 2^2+b^2&=3^2\\ \Rightarrow b&=\sqrt{5}. \end{align}\]

Therefore the equation of the hyperbola is

\[\frac{(x-4)^2}{4}-\frac{(y-3)^2}{5}=1.\ _\square\]

The figure below depicts the graph of the hyperbola: \(\frac{x^2}{16}-\frac{y^2}{9}=-1,\) where \(F\) and \(F'\) are its two foci. If the length of \(\overline{FP}\) is 4, what is the perimeter of \(\triangle FPF'?\)

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From the equation we have \(a=4\) and \(b=3.\) Since the hyperbola's axis is the \(y\)-axis, the absolute value of the difference of the distances from the two foci is \(2b=6.\) Since \(a^2+b^2=16+9=5^2,\) the coordinates of the foci are \(F=(0,5)\) and \(F'=(0,-5).\) We are given \(\lvert\overline{FP}\rvert=4,\) so \(\lvert\overline{F'P}\rvert=4+2b=10.\) Therefore the perimeter of \(\triangle FPF'\) is

\[\lvert\overline{FF'}\rvert+\lvert\overline{FP}\rvert+\lvert\overline{F'P}\rvert=10+4+10=24.\ _\square\]

What are the equations of the asymptotes of the hyperbola

\[\frac{x^2}{4}-\frac{y^2}{9}=-1?\]

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For a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm1,\) the equation of its asymptotes is given by the formula

\[y=\pm\frac{b}{a}x.\]

Therefore the answer is

\[y=\pm\frac{3}{2}x.\ _\square\]