lf the eccentricity of the hyperbola `x^2-y^2(sec)alpha=5` is `sqrt3` times the eccentricity of the ellipse `x^2(sec)^2alpha+y^2=25,` then a value of

Question

lf the eccentricity of the hyperbola `x^2-y^2(sec)alpha=5` is `sqrt3` times the eccentricity of the ellipse `x^2(sec)^2alpha+y^2=25,` then a value of

lf the eccentricity of the hyperbola `x^2-y^2(sec)alpha=5` is `sqrt3` times the eccentricity of the ellipse `x^2(sec)^2alpha+y^2=25,` then a value of `alpha` is : (a) `pi/6` (b) `pi/4` (c) `pi/3` (d) `pi/2`
A. `pi//6`
B. `pi//4`
C. `pi//3`
D. `pi//2`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - B
For the hyperbola
`(x^(2))/(5)-(y^(2))/(5 cos^(2)alpha)=1`
we have
`e_(1)^(2)=1+(b^(2))/(a^(2))=1+(5 cos^(2)alpha)/(5)=1+cos^(2)alpha`
For the ellipse
`(x^(2))/(25 cos^(2)alpha)+(y^(2))/(25)=1`
we have
`e_(2)^(2)=1-(25 cos^(2)alpha)/(25)=sin^(2)alpha`
Given that
`e_(1)=sqrt3e_(2)`
`therefore" "e_(1)^(2)=3e_(2)^(2)`
`"or "1+cos^(2)alpha=3 sin^(2)alpha`
`"or "2=4 sin^(2)alpha`
`"or "sin alpha=(1)/(sqrt2)`