Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \(\dfrac{y^2}{9}-\dfrac{x^2}{27}=1\)
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \(\dfrac{y^2}{9}-\dfrac{x^2}{27}=1\)
Solution and Explanation
The given equation is \(\dfrac{y^2}{9} – \dfrac{x^2}{27} = 1\)
On comparing this equation with the standard equation of hyperbola i.e., \(\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1\)
We obtain \(a = 3 \) and \(b = √27\).
We know that
\(a^2 + b^2 = c^2 .\)
\(∴ c^2 = 3^2 + (√27)^2\)
\(= 9 + 27\)
\(c^2 = 36\)
\(c = √36\)
\(= 6\)
Therefore, The coordinates of the foci are \((0, ±6).\)
The coordinates of the vertices are \(( 0,±3).\)
Eccentricity, \(e = \dfrac{c}{a} = \dfrac{6}{3} = 2\)
Length of the latus rectum \(= \dfrac{2b^2}{a} = \dfrac{(2 × 27)}{3} \)
\(= \dfrac{(54)}{3} =18\)
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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.
