Chapter 8.3, Problem 70E

Solution

The standard form for the equation of the hyperbola centered at the origin with the $x$ axis as the focal axis.

It has been determined that the standard form for the equation of the hyperbola centered at the origin with the $x$ axis as the focal axis, is $\frac{x^2}{1600}-\frac{y^2}{2000}=1$ .

Given:

A Cassegrain telescope has the dimensions shown in the figure:

  

Concept used:

Hyperbolas centered at the origin with the $x$ axis as the focal axis, has equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , foci at $\left(\pm c,0\right)$ and vertex at $\left(\pm a,0\right)$ , where $b^2=c^2-a^2$ .

Calculation:

Let the coordinate axes be superimposed such that the given hyperbola is centered at the origin with the $x$ axis as the focal axis.

According to the given figure, the distance between the two foci of the hyperbola, is $120$ cm.

Then, $2c=120$ .

Simplifying,

  $c=60$

According to the given figure, the distance between the two foci of the hyperbola is the sum of the distance between the left focus and the parabola and the distance between the parabola and the right focus.

It is given that the distance between the parabola and the right focus is $100$ cm and the distance between the two foci of the hyperbola is $120$ cm.

This implies that the distance between the left focus and the parabola, is $120-100=20$ cm.

According to the given figure, the distance between the left focus and the vertex of the hyperbola is the sum of the distance between the left focus and the parabola and the distance between the parabola and the hyperbola.

It is given that the distance between the parabola and the hyperbola is $80$ cm and it has been determined that the distance between the left focus and the parabola, is $20$ cm.

This implies that the distance between the left focus and the vertex of the hyperbola is $80+20=100$ cm.

Note that the referred vertex is not the nearer vertex to the referred focus for the hyperbola.

Then, the distance between them must be $a+c$ .

Now, according to the problem,

  $a+c=100$

Put $c=60$ in the above equation to get,

  $a=40$

Put $c=60$ and $a=40$ in $b^2=c^2-a^2$ to get,

  $\begin{array}{c}{b}^2={\left(60\right)}^2-{\left(40\right)}^2\\ =3600-1600\\ =2000\end{array}$

So, $b^2=2000$ .

Put $a=40$ and $b^2=2000$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to get,

  $\frac{x^2}{{\left(40\right)}^2}-\frac{y^2}{2000}=1$

Simplifying,

  $\frac{x^2}{1600}-\frac{y^2}{2000}=1$

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

Conclusion:

It has been determined that the standard form for the equation of the hyperbola centered at the origin with the $x$ axis as the focal axis, is $\frac{x^2}{1600}-\frac{y^2}{2000}=1$ .