Let the hyperbola
\(H:\frac{x^2}{a^2}−y^2=1\)
and the ellipse
\(E:3x^2+4y^2=12\)
be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of
\(12 (e^{2}_H+e^{2}_E)\) is equal to _____ .
Let the hyperbola
\(H:\frac{x^2}{a^2}−y^2=1\)
and the ellipse
\(E:3x^2+4y^2=12\)
be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of
\(12 (e^{2}_H+e^{2}_E)\) is equal to _____ .
Correct Answer: 42
Solution and Explanation
The correct answer is 42
\(∴ H:\frac{x^2}{a^2}−\frac{y^2}{1}=1\)
∴ Length of latus rectum
\(=\frac{2}{a}\)
\(E:3x^2+4y^2=12\)
Length of latus rectum
\(= \frac{6}{2} = 3\)
\(\because \frac{2}{a} = 3 ⇒ a = \frac{2}{3}\)
\(12 (e^{2}_H+e^{2}_E)\)
\( = 12(1+\frac{9}{4})+(1-\frac{3}{4})\)
\(= 42\)
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 \( OPQ \).
- 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 \( |\beta| + \frac{1}{|\alpha|} = \)
Questions Asked in JEE Main exam
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
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Consider the following two reactions A and B:
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
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Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ 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.
