Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \( \dfrac{x^2}{16}-\dfrac{y^2}{9}=2\)
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \( \dfrac{x^2}{16}-\dfrac{y^2}{9}=2\)
Solution and Explanation
The given equation is \( \dfrac{x^2}{16} – \dfrac{y^2}{9} = 1 \) or \(\dfrac{x^2}{4^2} –\dfrac{y^2}{3^2} = 1\)
On comparing this equation with the standard equation of hyperbola i.e., \( \dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1 \), we obtain \(a = 4\) and \(b = 3.\)
We know that
\(a ^2 + b ^2 = c^2 \)
\(∴ c^2 = 4^2 + 3^2\)
\(= √25\)
\(c = 5\)
Therefore, The coordinates of the foci are \(( ±5, 0).\)
The coordinates of the vertices are \(( ±4, 0).\)
Eccentricity, \(e = \dfrac{c}{a} = \dfrac{5}{4}\)
Length of the latus rectum \(= \dfrac{2b^2}{a} = \dfrac{(2 × 32)}{4} \)
\(= \dfrac{(2×9)}{4} \)
\(= \dfrac{18}{4}\)
\( = \dfrac{9}{2}\)
Top Questions on Hyperbola
- 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:
- If \( e_1 \) and \( e_2 \) are respectively the eccentricities of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and its conjugate hyperbola, then the line \( \frac{x}{2e_1} + \frac{y}{2e_2} = 1 \) touches the circle having center at the origin, then its radius is:
- If the eccentricity of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is \( \sec \alpha \), then the area of the triangle formed by the asymptotes of the hyperbola with any of its tangent is:
- If \( \tanh x = \text{sech } y = \frac{3}{5} \) and \( e^{x+y} \) is an integer, then \( e^{x+y} \) is:
Notes on Hyperbola
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.
