Question

Find eqn of ellipse whose focus is (4,9) and (-20,9) and eccentricity is 4.​

Answer 1

CORRECT QUESTION (as per user request)

Find eqn of ellipse whose focus is (4,9) and (-20,9) and eccentricity is 1/4.

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{ {(x+4)}^{2} }{2304} - \frac{ {(y-9)}^{2} }{2160} = 1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (4,9)\:and\:(-20,9)} \\ \\ \tt{ : \implies Eccentricity(e) = \frac{1}{4}} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2ae } \\ \\ \tt{ : \implies Foci = \sqrt{(4-(-20))^{2}+(9-9)^{2}}=24} \\ \\ \tt{ : \implies 2ae = 24} \\ \\ \tt{: \implies a \times \frac{1}{4} = 12} \\ \\ \green{ \tt{: \implies a= 48 }}\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} ( 1-{e}^{2} )} \\ \\ \tt{: \implies {b}^{2} = {48}^{2} (1-\frac{1}{16})} \\ \\ \green{ \tt{ : \implies {b}^{2} = 2304\times \frac{15}{16}} }\\ \\ \tt{: \implies {b}^{2} = 2160} \\\\ \tt{:\implies Co-ordinate\:of\:centre=\frac{4-20}{2},\frac{9+9}{2}}\\\\ \tt{:\implies Co-ordinate\:of\:centre=(-4,9) }\\\\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies \frac{ {(x+4)}^{2} }{2304} - \frac{ {(y-9)}^{2} }{2160} = 1}}$