The normal to the hyperbola
\(\frac{x²}{a²} - \frac{y²}{9} = 1\)
at the point (8, 3√3) on it passes through the point:
The normal to the hyperbola
\(\frac{x²}{a²} - \frac{y²}{9} = 1\)
at the point (8, 3√3) on it passes through the point:
\((15, -2\sqrt3)\)
\((9, 2\sqrt3)\)
\(( -1, 9\sqrt3)\)
\(( -1, 6\sqrt3)\)
The Correct Option is C
Solution and Explanation
The correct answer is (C) : \((-1,9\sqrt3)\)
Given hyperbola :
\(\frac{x²}{a²} - \frac{y²}{9} = 1\)
∵ It passes through
\((8, 3\sqrt3)\)
\(∵ \frac{64}{a²} - \frac{27}{9} = 1 ⇒ a² = 16\)
Now , equation of normal to hyperbola
\(\frac{16x}{8} + \frac{9y}{3\sqrt3} = 16 + 9\)
\(⇒ 2x + \sqrt3y = 25 ...... (i)\)
\((-1 , 9\sqrt3)\) satisfies (i)
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:
Questions Asked in JEE Main exam
- Consider ‘n’ is the number of lone pair of electrons present in the equatorial position of the most stable structure of \( \text{ClF}_3 \). The ions from the following with ‘n’ number of unpaired electrons are:
A. \( \text{V}^{3+} \)
B. \( \text{Ti}^{3+} \)
C. \( \text{Cu}^{2+} \)
D. \( \text{Ni}^{2+} \)
E. \( \text{Ti}^{2+} \)
Choose the correct answer from the options given below:- JEE Main - 2025
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Find the equivalent capacitance between A and B, where \( C = 16 \, \mu F \).
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If the equation of the parabola with vertex \( \left( \frac{3}{2}, 3 \right) \) and the directrix \( x + 2y = 0 \) is \[ ax^2 + b y^2 - cxy - 30x - 60y + 225 = 0, \text{ then } \alpha + \beta + \gamma \text{ is equal to:} \]
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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.
