Question:
The eccentricity of the hyperbola with latus rectum $ 12 $ and semi-conjugate axis $ 2\sqrt{3} $ , is
The eccentricity of the hyperbola with latus rectum $ 12 $ and semi-conjugate axis $ 2\sqrt{3} $ , is
Updated On: Jun 14, 2022
- $ 3 $
- $ \sqrt{\frac{3}{2}} $
- $ 2\sqrt{3} $
- 2
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Let equation of hyperbola be.
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Length of latusrectum, $ \frac{2b^2}{a} = 12$
and length of semi-coqjugate axis, $ b = 2\sqrt{3}$
$\therefore \frac{2(2\sqrt{3})^2}{a} = 12$
$\Rightarrow a = 2$
$\therefore$ Eccentricity$ e = \sqrt{1+ \frac{b^2}{a^2}}$
$ = \sqrt{1+ \frac{\left(2\sqrt{3}\right)^{2}}{\left(2\right)^{2}}}$
$= \sqrt{1 + \frac{12}{4}} =2$
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Length of latusrectum, $ \frac{2b^2}{a} = 12$
and length of semi-coqjugate axis, $ b = 2\sqrt{3}$
$\therefore \frac{2(2\sqrt{3})^2}{a} = 12$
$\Rightarrow a = 2$
$\therefore$ Eccentricity$ e = \sqrt{1+ \frac{b^2}{a^2}}$
$ = \sqrt{1+ \frac{\left(2\sqrt{3}\right)^{2}}{\left(2\right)^{2}}}$
$= \sqrt{1 + \frac{12}{4}} =2$
Was this answer helpful?
0
0
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:
- The foci of the hyperbola \[ 4x^2 - 9y^2 - 1 = 0 \] are:
- If \( e_1 \) and \( e_2 \) are respectively the eccentricities of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and its conjugate hyperbola, then the line \( \frac{x}{2e_1} + \frac{y}{2e_2} = 1 \) touches the circle having center at the origin, then its radius is:
- If the eccentricity of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is \( \sec \alpha \), then the area of the triangle formed by the asymptotes of the hyperbola with any of its tangent is:
- If \( \tanh x = \text{sech } y = \frac{3}{5} \) and \( e^{x+y} \) is an integer, then \( e^{x+y} \) is:
View More Questions
Questions Asked in AMUEEE exam
- The ratio of the half-life time $ (t_{1/2}) $ , to the three quarter life-time, $ (t_{3/4}) $ , for a reaction that is second order
- AMUEEE - 2018
- Order of Reaction
- $ 0.002\, M $ solution of a weak acid has an equivalent conductance $ (\wedge) 60\, ohm^{-1}\, cm^{2}\, eq^{-1} $ . What will be the $ pH $ ? (Given : $ \wedge = 400 \, ohm^{-1} \, cm^{2} \, eq^{-1} $ ]
- AMUEEE - 2018
- Electrochemistry
- The electrode potential, $ E^{\circ} $ , for the reduction of $ MnO^{-}_{4} $ to $ Mn^{2+} $ in acidic medium is $ +1.51\, V $ . Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for $ Zn^{2+}, Ag^{+} $ , and $ Au^{+} $ are given as $ Zn^{2+} (aq) +2e \to Zn(s), E^{\circ} = -0.762\, V $ $ Ag+ (aq) +e {\rightleftharpoons} Ag (s), E^{\circ}=+0.80 \,V $ $ Au^{+} (aq) +e {\rightleftharpoons} Au(s), E^{\circ}+1.69\,V $
- AMUEEE - 2018
- Electrochemistry
- Three particles, two with masses $ m $ and one with mass $ M $ , might be arranged in any of the four configurations shown below. Rank the configurations according to the magnitude of the gravitational force on $ M $ , least to greatest
- AMUEEE - 2018
- Gravitation
- The $ EAN $ value $ y \left[Ti\left(\sigma-C_{6}H_{5}\right)_{2}\left(\pi-C_{5}H_{5}\right)_{2}\right]^{0} $ is
- AMUEEE - 2018
- coordination compounds
View More Questions
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.
