Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)
Solution and Explanation
Vertices (±2, 0), foci (±3, 0)
Here, the vertices 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 vertices are (±2, 0), a = 2.
Since the foci are (±3, 0), c = 3.
We know that \(a ^2 + b^ 2 = c^ 2.\)
\(∴ 2^2 + b^2 = 3^2\)
\(b^2 = 9 – 4 = 5\)
∴ The equation of the hyperbola is \(\frac{x^2}{4} –\frac{ y^2}{5} = 1\)
Top Questions on Hyperbola
- If the normal drawn to the hyperbola \( xy = 16 \) at (8, 2) meets the hyperbola again at a point \((\alpha, \beta)\), then \( |\beta| + \frac{1}{|\alpha|} = \)
- If \( 3\sqrt{2}x - 4y = 12 \) is a tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(\frac{5}{4}\) is its eccentricity, then \( a^2 - b^2 = \)
- Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $ 2a $ and $ 2b $, respectively, and one focus and the corresponding directrix of this hyperbola be $ (-5, 0) $ and $ 5x + 9 = 0 $, respectively. If the product of the focal distances of a point $ (\alpha, 2\sqrt{5}) $ on the hyperbola is $ p $, then $ 4p $ is equal to:
- Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is:
- The foci of the hyperbola \[ 4x^2 - 9y^2 - 1 = 0 \] are:
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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.
