Question

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates if its vertices are $(0,\pm 7)$ and $e=\frac{4}{3}.$

• A. $\frac{9x^2}{343}+\frac{y^2}{49}=1$
• B. None of the above
• C. $\frac{9x^2}{343}-\frac{y^2}{49}=-1$
• D. $\frac{9x^2}{343}-\frac{y^2}{49}=1$

Answer 1

The correct option is C $\frac{9x^2}{343}-\frac{y^2}{49}=-1$

Since the vertices of the required hyperbola lie on $y$-axis. So, let its equation be

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1...\left(i\right)$

The coordinates of vertices of this hyperbola are $(0,\pm b).$ So, $b=7$

Now, $a^2=b^2(e^2-1)\Rightarrow a^2=49(\frac{16}{9}-1)\Rightarrow a^2=49\times \frac{7}{9}=\frac{343}{9}$

Substituting the values of $a^2$ and $b^2$ in (i), we get

$\frac{9x^2}{343}-\frac{y^2}{49}=-1$ as the equation of the desired hyperbola.