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 lengths of the transverse and conjugate axes of a hyperbola in standard form be $ 2a $ and $ 2b $, respectively, and one focus and the corresponding directrix of this hyperbola be $ (-5, 0) $ and $ 5x + 9 = 0 $, respectively. If the product of the focal distances of a point $ (\alpha, 2\sqrt{5}) $ on the hyperbola is $ p $, then $ 4p $ is equal to:
- If the normal drawn to the hyperbola \( xy = 16 \) at (8, 2) meets the hyperbola again at a point \((\alpha, \beta)\), then \( |\beta| + \frac{1}{|\alpha|} = \)
- If \( 3\sqrt{2}x - 4y = 12 \) is a tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(\frac{5}{4}\) is its eccentricity, then \( a^2 - b^2 = \)
- 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:
Questions Asked in JEE Main exam
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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
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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.
