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
- If the line \( ax + 2y = 1 \), where \( a \in \mathbb{R} \), does not meet the hyperbola \( x^2 - 9y^2 = 9 \), then a possible value of \( a \) is:
- Let \( PQ \) be a chord of the hyperbola \[ \frac{x^2}{4} - \frac{y^2}{b^2} = 1, \] perpendicular to the \(x\)-axis such that \( OPQ \) is an equilateral triangle, \( O \) being the centre of the hyperbola. If the eccentricity of the hyperbola is \( \sqrt{3} \), then find the area of the triangle \( OPQ \).
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- 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 \( |\beta| + \frac{1}{|\alpha|} = \)
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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.
