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 \(\frac{x^2}{4} –\frac{ y^2}{5} = 1\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
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.