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:
- 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 \( a^2 - b^2 = \)
- 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:
Questions Asked in JEE Main exam
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
- JEE Main - 2025
- Conic sections
- Let \( \alpha_1 \) and \( \beta_1 \) be the distinct roots of \( 2x^2 + (\cos\theta)x - 1 = 0, \ \theta \in (0, 2\pi) \). If \( m \) and \( M \) are the minimum and the maximum values of \( \alpha_1 + \beta_1 \), then \( 16(M + m) \) equals:
- JEE Main - 2025
- Maxima and Minima
- Let the line \( x + y = 1 \) meet the circle \( x^2 + y^2 = 4 \) at the points A and B. If the line perpendicular to AB and passing through the midpoint of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to:
- JEE Main - 2025
- Coordinate Geometry
- The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is:
- JEE Main - 2025
- permutations and combinations
- Two identical symmetric double convex lenses of focal length \( f \) are cut into two equal parts \( L_1, L_2 \) by the AB plane and \( L_3, L_4 \) by the XY plane as shown in the figure respectively. The ratio of focal lengths of lenses \( L_1 \) and \( L_3 \) is:
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.
