Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola
\(\frac{x^2}{a^2}−\frac{y^2}{b^2}=1\)
Let e′ and l′ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If
\(e^2=\frac{11}{14}l\) and \((e^′)^2=\frac{11}{8}l^′\)
then the value of 77a + 44b is equal to :
Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola
\(\frac{x^2}{a^2}−\frac{y^2}{b^2}=1\)
Let e′ and l′ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If
\(e^2=\frac{11}{14}l\) and \((e^′)^2=\frac{11}{8}l^′\)
then the value of 77a + 44b is equal to :
100
110
120
130
The Correct Option is D
Solution and Explanation
The correct answer is (D) : 130
\(H : \frac{x^2}{a^2}−\frac{y^2}{b^2}=1\)
then
\(e^2=\frac{11}{14}l\)
(I be the length of LR)
\(⇒a^2+b^2=\frac{11}{7}b^2a…(i)\)
and
\(e^{′^2}=\frac{11}{8}l^′\)
(I′ be the length of LR of conjugate hyperbola)
\(⇒a^2+b^2=\frac{11}{4}a^2b…(ii)\)
By (i) and (ii)
7a = 4b
then by (i)
\(\frac{16}{49}b^2+b^2=\frac{11}{7}b^2⋅\frac{4b}{7}\)
⇒ 44b = 65 and 77a = 65
Therefore , 77a + 44b = 130
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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.
