Chapter 8.3, Problem 36E

Solution

An equation in standard form for the hyperbola that satisfies the given conditions.

It has been determined that the equation of the given hyperbola in standard form is $\frac{4{\left(y+\frac{11}{2}\right)}^2}{49}-\frac{{\left(x+3\right)}^2}{18}=1$ .

Given:

Transverse axis end points $\left(-3,-9\right)$ and $\left(-3,-2\right)$ ; foci $\left(-3,-11\right)$ and $\left(-3,0\right)$ .

Concept used:

The equation of a hyperbola in standard form with focal axis $y=k$ , foci at $\left(h\pm c,k\right)$ , vertices at $\left(h\pm a,k\right)$ , semi-transverse axis of length $a$ , semi-conjugate axis of length $b$ and asymptotes $y=\pm \frac{b}{a}\left(x-h\right)+k$ ; is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , where $c=\sqrt{a^2+b^2}$ .

The equation of a hyperbola in standard form with focal axis $x=h$ , foci at $\left(h,k\pm c\right)$ , vertices at $\left(h,k\pm a\right)$ , semi-transverse axis of length $a$ , semi-conjugate axis of length $b$ and asymptotes $y=\pm \frac{a}{b}\left(x-h\right)+k$ ; is $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ , where $c=\sqrt{a^2+b^2}$ .

Calculation:

It is given that the transverse axis end points of the given hyperbola are $\left(-3,-9\right)$ and $\left(-3,-2\right)$ .

That is, the vertices of the given hyperbola are $\left(-3,-9\right)$ and $\left(-3,-2\right)$ .

Comparing these vertices with $\left(h,k\pm a\right)$ , it follows that:

  $h=-3$ , $k+a=-2$ and $k-a=-9$ .

Adding the equations; $k+a=-2$ and $k-a=-9$ , it follows that,

  $2k=-11$

Simplifying,

  $k=-\frac{11}{2}$

Put $k=-\frac{11}{2}$ in $k+a=-2$ to get $a=\frac{7}{2}$ .

It is also given that foci of the given hyperbola are $\left(-3,-11\right)$ and $\left(-3,0\right)$ .

Comparing these foci with $\left(h,k\pm c\right)$ , it follows that:

  $h=-3$ , $k+c=0$ and $k-c=-11$ .

Put $k=-\frac{11}{2}$ in $k+c=0$ to get $c=\frac{11}{2}$ .

Now, put $c=\frac{11}{2}$ and $a=\frac{7}{2}$ in $c=\sqrt{a^2+b^2}$ to get,

  $\begin{array}{c}\frac{11}{2}=\sqrt{{\left(\frac{7}{2}\right)}^2+b^2}\\ {\left(\frac{11}{2}\right)}^2={\left(\frac{7}{2}\right)}^2+b^2\\ b^2={\left(\frac{11}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2\\ b=\pm \sqrt{{\left(\frac{11}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2}\end{array}$

Simplifying,

  $\begin{array}{c}b=\pm \sqrt{{\left(\frac{11}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2}\\ =\pm \frac{1}{2}\sqrt{{\left(11\right)}^2-{\left(7\right)}^2}\\ =\pm \frac{1}{2}\sqrt{\left(11-7\right)\left(11+7\right)}\\ =\pm \frac{1}{2}\sqrt{\left(4\right)\left(18\right)}\end{array}$

Solving,

  $\begin{array}{c}b=\pm \frac{1}{2}\left(2\right)\left(3\right)\sqrt{2}\\ =\pm 3\sqrt{2}\end{array}$

So, $b=\pm 3\sqrt{2}$ .

Put $h=-3$ , $k=-\frac{11}{2}$ , $a=\frac{7}{2}$ and $b=\pm 3\sqrt{2}$ in $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ to get,

  $\frac{{\left(y-\left(-\frac{11}{2}\right)\right)}^2}{{\left(\frac{7}{2}\right)}^2}-\frac{{\left(x-\left(-3\right)\right)}^2}{{\left(\pm 3\sqrt{2}\right)}^2}=1$

Simplifying,

  $\frac{4{\left(y+\frac{11}{2}\right)}^2}{49}-\frac{{\left(x+3\right)}^2}{18}=1$

This is the equation of the given hyperbola in standard form.

Conclusion:

It has been determined that the equation of the given hyperbola in standard form is $\frac{4{\left(y+\frac{11}{2}\right)}^2}{49}-\frac{{\left(x+3\right)}^2}{18}=1$ .