The length of the transverse axis of a hyperbola, 2 cos0. the foci of the hyperbola are the same as that of the ellips `9x^(2)+16y^(2)=144. the equation of the hypperbola is

Options

A. `x^(2)/(cos^2alpha)-y^(2)/(7-cos^2alpha)`=1
B. `x^(2)/(cos^2alpha)-y^(2)/(7+cos^2alpha)`=1
C. `x^(2)/(+1cos^2alpha)-y^(2)/(7-cos^2alpha)`=1
D. `x^(2)/(+1cos^2alpha)-y^(2)/(7+cos^2alpha)`=1

Answer

A

Explanation

Let equation of hyperbola be `(x^2)/(a1^2)-(y^2)/(b1^(2))=1`
given, `2a_1=2 cosalpha rArra_1=cos alpha`
` also, given equation of ellipes is `(x^2)/(16)+y^(2)/9=1`
`here, e= `sqrt(1-b^(2)/a^(2)=)`sqrt(1-9/16=)`sqrt(7/4`
` rArr+-a_1e_1=+-ae`
`cos alpha .e_1=4. sqrt7/4` rArrcos alpha .e_1= sqrt7`
`rArr cos alpha sqrt `
` (1+b_1^2/cos^2alpha=sqrt7`
`rArr cos^(2)+b_1^(2)=7`
`rArrb_1^(2)=7-cos^(2)alpha`
` `therefore`the equation of hyperbola is `(x^(2))/(cos^(2))-y^(2)/(7-cos^(2)alpha)`=1`