Question

Find the equation of hyperbola in standard with foci (2,5) and (2,9) with eccentricity 4

Answer 1

For hyperbola we know,
the difference of focal distance of any point on the hyperbola is equal to Length of transverse axis of the hyperbola .

Let P (x, y) is the point on hyperbola,
Let S(2, 9) and S'(2, 5)
Length of transverse axis = (9 - 5)/4 = 1
for hyperbola ,
SP - S'P = length of transverse axis
√{(x -2)² +(y-9)²} - √{x-2)²+(y-5)²}= 1
√{(x-2)²+ (y-9)²} = 1 + √{(x-2)² +(y-5)² }
take square both sides,
(x-2)² + (y-9)² = 1 + (x-2)² + (y-5)² +2√{(x-2)²+(y-5)²}
(y-9)²-(y-5)² -1 = 2√{(x-2)²+(y-5)²}
-18y +81 +10y -25 -1 = 2√{(x-2)²+(y-5)²}
(-8y +55)/2 = √{(x-2)²+(y-5)²}
take square both sides,
(-8y +55)²= 4(x-2)² + 4(y-5)² is the equation of hyperbola

Answer 2

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=4(x-2)^{2}-30(y-7)^{2}=-15}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (2,5)\:and\:(2,9)} \\ \\ \tt{ : \implies Eccentricity(e) = 4} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: -\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2be } \\ \\ \tt{ : \implies Foci \ = \sqrt{(2-2)^{2}+(9-5)^{2}}=4} \\ \\ \tt{ : \implies 2be = 4} \\ \\ \tt{: \implies b \times {4} = 2} \\ \\ \green{ \tt{: \implies b = \frac{1}{2}} }\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {a}^{2} = {b}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {a}^{2} = {(\frac{1}{2})}^{2} ( 4^{2} - 1)} \\ \\ \green{ \tt{ : \implies {a}^{2} = \frac{1}{4}\times 15} }\\ \\ \tt{: \implies {a}^{2} = \frac{15}{4}} \\\\ \tt{:\implies Co-ordinate\:of\:centre=\frac{2+2}{2},\frac{5+9}{2}}\\\\ \tt{:\implies Co-ordinate\:of\:centre=(2,7) }\\\\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies -\frac{ {(x-2)}^{2} }{\frac{15}{4}} + \frac{ {(y-7)}^{2} }{\frac{1}{2}} = 1}} \\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: 4{(x-2)}^{2} - 30{(y-7)}^{2} = -15}}$