MCQ
The distance between the directrices of the hyperbola $\text{x}=8\sec\theta,\text{y}=8,$ is
- ✓$8\sqrt2$
- B$16\sqrt2$
- C$4\sqrt2$
- D$6\sqrt2$
✓
Answer
Correct option: A.
$8\sqrt2$
We have:
$\text{x}=8\sec\theta,\text{y}=8\tan\theta$
On squaring and subtracting:
$\text{x}^2-\text{y}^2=8\sec^2\theta-8\tan^2\theta$
$\Rightarrow\text{x}^2-\text{y}^2=8$
$\Rightarrow\frac{\text{x}^2}{8}-\frac{\text{y}^2}{8}=1$
$\therefore\text{a}=\text{b}=\text{c}$
Distance between the directrices of the hyperbola $=\frac{2\text{a}^2}{\sqrt{\text{a}^2+\text{b}^2}}$
Distance between the directrices $=\frac{2\times64}{\sqrt{64+64}}$
$=\frac{128}{8\sqrt2}$
$=\frac{16}{\sqrt2}$
$=8\sqrt2$
$\text{x}=8\sec\theta,\text{y}=8\tan\theta$
On squaring and subtracting:
$\text{x}^2-\text{y}^2=8\sec^2\theta-8\tan^2\theta$
$\Rightarrow\text{x}^2-\text{y}^2=8$
$\Rightarrow\frac{\text{x}^2}{8}-\frac{\text{y}^2}{8}=1$
$\therefore\text{a}=\text{b}=\text{c}$
Distance between the directrices of the hyperbola $=\frac{2\text{a}^2}{\sqrt{\text{a}^2+\text{b}^2}}$
Distance between the directrices $=\frac{2\times64}{\sqrt{64+64}}$
$=\frac{128}{8\sqrt2}$
$=\frac{16}{\sqrt2}$
$=8\sqrt2$
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