Question

Find the equation of the hyperbola whose vertices are (±3, 0) and the foci are (±4, 0)
1. \(\rm \frac{x^2}{3}- \frac{y^2}{7}=1\)
2. \(\rm \frac{x^2}{16}- \frac{y^2}{9}=1\)
3. \(\rm \frac{x^2}{7}- \frac{y^2}{3}=1\)
4. \(\rm \frac{x^2}{9}- \frac{y^2}{7}=1\)

Answer 1

Correct Answer - Option 4 : \(\rm \frac{x^2}{9}- \frac{y^2}{7}=1\)

Concept:

Equation of Hyperbola: \(\rm \frac{x^2}{a^2}- \frac{y^2}{b^2}=1\)  

Eccentricity: \(\rm e= \rm\sqrt{1+ \frac{b^2}{a^2}} \)

Vertices =  (±a, 0)

Focus = (±ae, 0)

 

Calculation:

The vertices of the given hyperbola are (±3, 0) which are in the form of  (±a, 0)

∴ a = 3

Also, the foci are  (±4, 0), which is in the form of (±ae, 0)

∴ ae = 4

⇒ e = 4/3                   .....(∵ a = 3)

Now, we know, \(\rm e= \rm\sqrt{1+ \frac{b^2}{a^2}} \)

\(\rm \Rightarrow \frac43=\sqrt{1+\frac{b^2}{3^2}}\)

\(\rm \Rightarrow \frac{16}{9}={\frac{9+b^2}{9}}\)        .....(squaring both sides)

\(\rm \Rightarrow 16={9+b^2}\)

⇒ b² = 7

∴ Equation of hyperbola = \(\rm \frac{x^2}{9}- \frac{y^2}{7}=1\)

Hence, option (4) is correct.