Chapter 8.3, Problem 53E

Solution

To Prove: The equation of the given hyperbola is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , where $c>a>0$ and $b^2=c^2-a^2$ .

It has been shown that the equation of the given hyperbola is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , where $c>a>0$ and $b^2=c^2-a^2$ .

Given:

A hyperbola with center $\left(0,0\right)$ , foci $\left(0,\pm c\right)$ and semi-transverse axis $a$ .

Concept used:

The difference of the distances from the foci to each point on the hyperbola is constant.

Calculation:

It is given that the hyperbola has center $\left(0,0\right)$ , foci $\left(0,\pm c\right)$ and semi-transverse axis $a$ .

Then, the vertices of the hyperbola must be $\left(0,\pm a\right)$ .

Consider the vertex $\left(0,a\right)$ . The distance between this vertex and the focus $\left(0,c\right)$ is $c-a$ since $c>a>0$ and the distance between this vertex and the focus $\left(0,-c\right)$ is $c+a$ .

So, the difference of the distances from the foci to this vertex is $\left(c+a\right)-\left(c-a\right)=2a$ .

It is given that the difference of the distances from the foci to each point on the hyperbola is constant.

Since the vertex is also a point on the hyperbola, it follows that the difference of the distances from the foci to each point on the hyperbola is $2a$ .

Now, depending on if the point is on the upper branch of the hyperbola or the lower branch of the hyperbola, the distance between that point and the upper focus is lesser than the distance between that point and the lower focus and vice versa.

Hence, it can be said that if $P\left(x,y\right)$ is any point on the hyperbola, $F_1\left(0,c\right)$ is the upper focus and $F_2\left(0,-c\right)$ is the lower focus, then $P{F}_1-P{F}_2=\pm 2a$ .

Now, applying distance formula,

  $P{F}_1=\sqrt{{\left(x-0\right)}^2+{\left(y-c\right)}^2}=\sqrt{x^2+{\left(y-c\right)}^2}$ ,

  $P{F}_2=\sqrt{{\left(x-0\right)}^2+{\left(y+c\right)}^2}=\sqrt{x^2+{\left(y+c\right)}^2}$

Put these values in $P{F}_1-P{F}_2=\pm 2a$ to get,

  $\sqrt{x^2+{\left(y-c\right)}^2}-\sqrt{x^2+{\left(y+c\right)}^2}=\pm 2a$

Simplifying,

  $\sqrt{x^2+{\left(y-c\right)}^2}=\pm 2a+\sqrt{x^2+{\left(y+c\right)}^2}$

Squaring both sides,

  $x^2+{\left(y-c\right)}^2=4a^2+x^2+{\left(y+c\right)}^2+2\left(\pm 2a\right)\sqrt{x^2+{\left(y+c\right)}^2}$

Simplifying,

  ${\left(y-c\right)}^2-{\left(y+c\right)}^2=4a^2\pm 4a\sqrt{x^2+{\left(y+c\right)}^2}$

On further simplification,

  $-4cy=4a\left(a\pm \sqrt{x^2+{\left(y+c\right)}^2}\right)$

Continuing simplification,

  $\begin{array}{l}\pm \sqrt{x^2+{\left(y+c\right)}^2}=-\frac{cy}{a}-a\\ \pm \sqrt{x^2+{\left(y+c\right)}^2}=-\left(\frac{cy+a^2}{a}\right)\end{array}$

Squaring both sides,

  $\begin{array}{c}{x}^2+{\left(y+c\right)}^2={\left(\frac{cy+a^2}{a}\right)}^2\\ a^2\left(x^2+{\left(y+c\right)}^2\right)={\left(cy+a^2\right)}^2\\ a^2\left(x^2+y^2+c^2+2cy\right)=c^2{y}^2+a^4+2a^{2}cy\\ a^2{x}^2+a^2{y}^2+a^2{c}^2+2a^{2}cy=c^2{y}^2+a^4+2a^{2}cy\end{array}$

Simplifying,

  $a^2{x}^2+a^2{y}^2+a^2{c}^2=c^2{y}^2+a^4$

Put $c^2=a^2+b^2$ in the above expression to get,

  $\begin{array}{c}{a}^2{x}^2+a^2{y}^2+a^2\left(a^2+b^2\right)=\left(a^2+b^2\right)y^2+a^4\\ a^2{x}^2+a^2{y}^2+a^4+a^2{b}^2=a^2{y}^2+b^2{y}^2+a^4\\ a^2{x}^2+a^2{b}^2=b^2{y}^2\\ b^2{y}^2-a^2{x}^2=a^2{b}^2\end{array}$

Solving,

  $\begin{array}{c}\frac{b^2{y}^2-a^2{x}^2}{a^2{b}^2}=1\\ \frac{b^2{y}^2}{a^2{b}^2}-\frac{a^2{x}^2}{a^2{b}^2}=1\\ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\end{array}$

This shows that the equation of the given hyperbola is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ .

Conclusion:

It has been shown that the equation of the given hyperbola is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , where $c>a>0$ and $b^2=c^2-a^2$ .