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# The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is - Mathematics

MCQ

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is

#### Options

• $8\sqrt{2}$

• $16\sqrt{2}$

• $4\sqrt{2}$

• $6\sqrt{2}$

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#### Solution

$8\sqrt{2}$

We have: $6\sqrt{2}$ $x = 8\sec\theta, y = 8\tan\theta$

On squaring and subtracting:

$x^2 - y^2 = 8 \sec^2 \theta - 8 \tan^2 \theta$

$\Rightarrow x^2 - y^2 = 8$

$\Rightarrow \frac{x^2}{8} - \frac{y^2}{8} = 1$

a = b = 8

Distance between the directrices of the hyperbola = $\frac{2 a^2}{\sqrt{a^2 + b^2}}$

Distance between the directrices = $\frac{2 \times 64}{\sqrt{64 + 64}}$

$= \frac{128}{8\sqrt{2}}$

$= \frac{16}{\sqrt{2}}$

$= 8\sqrt{2}$

Concept: Introduction of Hyperbola
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Q 3 | Page 19
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