Question

please solve this question​

Answer 1

Answer:-

Given ellipse,

$\sf \frac{ {x}^{2} }{25} + \frac{{y}^{2}}{9} = 1$

Comparing with $\sf \frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1, we get, </p><p>[tex] \sf {a}^{2} = 25 \: and \: {b}^{2} = 9$

Then,

Eccentricity e$\sf = \sqrt{1 - \frac{ {b}^{2} }{ {a}^{2} } }$

$\sf = \sqrt{1 - \frac{9}{25} }$

$\sf = \sqrt{ \frac{16}{25} }$

$\sf = \frac{4}{5}$

Foci of the ellipse are (±4, 0).

So, the coordinates of the foci of the hyperbola are (±4, 0)

Let e' be the eccentricity of the required hyperbola and it's equation be

$\sf \frac{ {x}^{2} }{{A}^{2} } - \frac{ {y}^{2} }{B^{2} } = 1 \: ....(i)$

Therefore, the coordinates of the foci are (±Ae', 0)

$\sf \implies Ae' = 4$

$\sf \implies A \times 2 = 4$

$\sf \implies A = 2$

Also,

$\sf {B}^{2} = A^{2}( {(e')}^{2} - 1)$

$\sf = 12$

Substituting the values of A and B in (i), we get,

$\sf \frac{ {x}^{2} }{4} - \frac{ {y}^{2} }{12} = 1$

$\sf \implies 3 {x}^{2} - {y}^{2} - 12 = 0$

Which is the required hyperbola.