Question

Fill in the blanks. (i) The length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ is . (ii) The equations of the hyperbola with vertices $\left(\pm 2, 0\right)$, foci $\left(\pm3, 0\right)$ is . (iii) If the distance between the foci of a hyperbola is $16$ and its eccentricity is $2$, then equation of the hyperbola is .

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

c

Answer 2

$\left(i\right)$ Given equation of hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ Now, $a^{2} = 16$ $\Rightarrow a=4$ and $b^{2}=9$ $\Rightarrow b=3$ Length of latus rectum $=\frac{2b^{2}}{a}=\frac{2\times9}{4}=\frac{9}{2}\cdot$ $\left(ii\right)$ Vertices are $\left(\pm2,0\right)$ which lie on $x$-axis. So the equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ Now, vertices are $\left(\pm2,0\right)$ $\Rightarrow a=2$ Foci are $\left(\pm3,0\right)$ $\Rightarrow ae=3$ $\Rightarrow e=\frac{3}{2}$ We know that $b=a\sqrt{e^{2}-1}$ $\Rightarrow b=2\sqrt{\frac{9}{4}}-1=2\frac{\sqrt{5}}{2}=\sqrt{5}$ Thus required equation of hyperbola is $\frac{x^{2}}{\left(2\right)^{2}}-\frac{y^{2}}{\left(\sqrt{5}\right)^{2}}=1$ $\Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$ $\left(iii\right)$ Since distance between the foci is $16$ $\Rightarrow 2ae - 16$ $\Rightarrow ae=8$ $\Rightarrow a=4 \, \left[\because\,e=2\right]$ Now, $b^{2} =a^{2} \left(e^{2}-1\right)=\left(4\right)^{2} \left[\left(2\right)^{2}-1\right]=16\times3=48$ Hence, the equation of hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{48}=1$ $\Rightarrow 3x^{2}-y^{2}=48$