Hyperbola Ques 15
- A hyperbola has its centre at the origin, passes through the point $(4,2)$ and has transverse axis of length 4 along the $X$-axis. Then the eccentricity of the hyperbola is
(2019 Main, 9 Jan II)
(a) 2
(b) $\frac{2}{\sqrt{3}}$
(c) $\frac{3}{2}$
(d) $\sqrt{3}$
Show Answer
Answer:
Correct Answer: 15.(b)
Solution:
- Equation of hyperbola is given by
$$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$
$\because$ Length of transverse axis $=2 a=4$
$$ \therefore \quad a=2 $$
Thus, $\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1$ is the equation of hyperbola
$\because$ It passes through $(4,2)$.
$\therefore \frac{16}{4}-\frac{4}{b^{2}}=1 \Rightarrow 4-\frac{4}{b^{2}}=1 \Rightarrow b^{2}=\frac{4}{3} \Rightarrow b=\frac{2}{\sqrt{3}}$
Now, eccentricity,
$$ e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{\frac{4}{3}}{4}}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} $$
BETA
