Advertisement Remove all ads

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

  • \[8\sqrt{2}\]

  • \[16\sqrt{2}\]

  • \[4\sqrt{2}\]

  • \[6\sqrt{2}\]

Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Q 3 | Page 19
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×