Find the equation of the hyperbola satisfying the give conditions: Foci (\(± 3\sqrt5, 0\)), the latus rectum is of length 8.
Solution and Explanation
Foci (\(± 3\sqrt5, 0\)), the latus rectum is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1. \)
Since the foci are (\(± 3\sqrt5, 0\)), \(c = ± 3\sqrt5 . \)
Length of latus rectum = 8
\(⇒\frac{ 2b^2}{a} = 8\)
\(⇒ 2b^2 = 8a\)
\(⇒ b^2 = \frac{8a}{2}\)
\(= 4a\)
We know that
a2 + b2 = c2
a2 + 4a = 45
a2 + 4a – 45 = 0
a2 + 9a – 5a – 45 = 0
(a + 9) (a -5) = 0
a = -9 or 5
Since a is non-negative, a = 5.
\(∴ b^2 = 4a= 4 \times 5= 20\)
Thus, the equation of the hyperbola is \(\frac{x^2}{25} –\frac{ y^2}{20} = 1\)
Top Questions on Hyperbola
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- Consider a hyperbola \( H \) having its centre at the origin and foci on the \( x \)-axis. Let \( C_1 \) be the circle touching the hyperbola \( H \) and having its centre at the origin. Let \( C_2 \) be the circle touching the hyperbola \( H \) at its vertex and having its centre at one of its foci. If the areas (in square units) of \( C_1 \) and \( C_2 \) are \( 36\pi \) and \( 4\pi \), respectively, then the length (in units) of the latus rectum of \( H \) is:
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
