Vertices (0, ±3), foci (0, ±5)
Here, the vertices are on the y-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 (0, ±3), a = 3.
Since the foci are (0, ±5), c = 5.
We know that \(a^2 + b^2 = c^2 .\)
\(∴ 3^2 + b^2 = 5^2\)
\(⇒ b^2 = 25 – 9 = 16\)
Thus, the equation of the hyperbola is\( \frac{y^2}{9} – \frac{x^2}{16} = 1\)
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.