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
\(= \frac{6}{2} = 3\)
\(\because \frac{2}{a} = 3 ⇒ a = \frac{2}{3}\)
\(12 (e^{2}_H+e^{2}_E)\)
\( = 12(1+\frac{9}{4})+(1-\frac{3}{4})\)
\(= 42\)
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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.
