Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

Find the Equation of the Hyperbola Whose Vertices Are at (± 6, 0) and One of the Directrices Is X = 4. - Mathematics

Answer in Brief

Find the equation of the hyperbola whose vertices are at (± 6, 0) and one of the directrices is x = 4.

Advertisement Remove all ads

Solution

The Vertices of the hyperbola are \[\left( \pm 6, 0 \right)\]

∴ \[a = 6\] 

⇒ a2 = 36
Now, x = 4

\[\frac{a}{e} = 4\]

\[ \Rightarrow e = \frac{3}{2} \left[ \because a = 6 \right]\]

Now,

\[\left( ae \right)^2 = a^2 + b^2 \]

\[ \Rightarrow \left( 6 \times \frac{3}{2} \right)^2 = 6^2 + b^2 \]

\[ \Rightarrow 81 - 36 = b^2 \]

\[ \Rightarrow b^2 = 45\]

Therefore, the equation of the hyperbola is \[\frac{x^2}{36} - \frac{y^2}{45} = 1\].

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

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 7.5 | Page 14
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×