Question:
Let P (4, 3) be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1.$ If the normal at P intersects the X-axis at (16, 0), then the eccentricity of the hyperbola is
Let P (4, 3) be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1.$ If the normal at P intersects the X-axis at (16, 0), then the eccentricity of the hyperbola is
Updated On: Mar 10, 2025
- $\frac{\sqrt{5}}{2}$
- $2$
- $\sqrt{2}$
- $\sqrt{3}$
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Shivam
The Correct Option is B
Solution and Explanation
Normal at P(4, 3)
$\frac{a^{2}x}{4} + \frac{b^{2}y}{3} = a^{2}+b^{2}$ through (16, 0)
$\Rightarrow 4a^{2} = a^{2} + b^{2}$
$\Rightarrow \frac{b^{2}}{a^{2}} = 3\,\therefore\, e = 2$
$\frac{a^{2}x}{4} + \frac{b^{2}y}{3} = a^{2}+b^{2}$ through (16, 0)
$\Rightarrow 4a^{2} = a^{2} + b^{2}$
$\Rightarrow \frac{b^{2}}{a^{2}} = 3\,\therefore\, e = 2$
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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.
