Question:
The equation of the hyperbola with vertices (3,0), (-3,0) and semi-latus rectum 4 is given by:
The equation of the hyperbola with vertices (3,0), (-3,0) and semi-latus rectum 4 is given by:
Updated On: May 12, 2022
- $4x^2 - 3y^2 + 36 = 0$
- $4x^2 - 3y^2 + 12 = 0$
- $4x^2 - 3y^2 - 36 = 0$
- $4x^2 -+3y^2 - 25 = 0$
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The Correct Option is C
Solution and Explanation
We have $a = 3$ and $\frac{b^{2}}{a} = 4 \Rightarrow b^{2} = 12$
Hence, the equation of the hyperbola is
$\frac{x^{2}}{9}-\frac{y^{2}}{12} = 1$
$\Rightarrow 4x^{2} - 3y^{2} = 36 \Rightarrow 4x^{2} - 3y^{2} - 36 = 0$
Hence, the equation of the hyperbola is
$\frac{x^{2}}{9}-\frac{y^{2}}{12} = 1$
$\Rightarrow 4x^{2} - 3y^{2} = 36 \Rightarrow 4x^{2} - 3y^{2} - 36 = 0$
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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.
