Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.
Solution and Explanation
Foci (±5, 0), the transverse axis is of length 8.
Here, the foci 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 foci are (±5, 0), c = 5.
Since the length of the transverse axis is 8,
2a = 8
\(⇒\) a = 4.
We know that
a2 + b2 = c2 .
∴ 42 + b2 = 52
\(⇒\) b2 = 25 – 16= 9
Thus, the equation of the hyperbola is \(\frac{x^2}{16} – \frac{y^2}{9} = 1\)
Top Questions on Hyperbola
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- Consider a hyperbola \( H \) having its centre at the origin and foci on the \( x \)-axis. Let \( C_1 \) be the circle touching the hyperbola \( H \) and having its centre at the origin. Let \( C_2 \) be the circle touching the hyperbola \( H \) at its vertex and having its centre at one of its foci. If the areas (in square units) of \( C_1 \) and \( C_2 \) are \( 36\pi \) and \( 4\pi \), respectively, then the length (in units) of the latus rectum of \( H \) is:
- Let H: $\frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$, $\alpha>0$ lies on H. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to:
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- The foci of a hyperbola are (±2,0) and its eccentricity is \(\frac{3}{2}\) . A tangent, perpendicular to the line 2x + 3y = 6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x- and y-axes are a and b respectively, then |6a| + |5b| is equal to_____.
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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.
