Chapter 9, Problem 43RE

To determine

Calculate the standard equation of the hyperbola.

Answer to Problem 43RE

The standard equation of the hyperbola is $\frac{x^2}{16}-\frac{y^2}{20}=1$ .

Explanation of Solution

Given:

It is given in the question that the $vertices:(\pm 4,0);Foci:(\pm 6,0)$ .Concept Used:

In this, use the concept of hyperbola and the equation of the parabola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and also use the formula of $b=\sqrt{c^2-a^2}$ .

Calculation: Since, the centre is at the origin. The distance between one of the foci and the center is $6$ .

So, $c=6$

Now, The vertices lie on the horizontal line, so the hyperbola has a horizontal transverse axis. The Distance between one of the vertices and the centre is $4$ ,

So, $a=4$

Now, use the formula of $b=\sqrt{c^2-a^2}$ ,

  $\begin{array}{l}b=\sqrt{c^2-a^2}\\ b=\sqrt{6^2-4^2}\\ b=\sqrt{36-16}\\ b=\sqrt{20}\\ b=2\sqrt{5}\end{array}$

So, the equation of Hyperbola is,

  $\begin{array}{l}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{4^2}-\frac{y^2}{{\sqrt{20}}^2}=1\\ \frac{x^2}{16}-\frac{y^2}{20}=1\end{array}$

Conclusion:

The equation is $\frac{x^2}{16}-\frac{y^2}{20}=1$ .