Question

Find the equation of the hyperbola whose foci are (8,3) and (0,3) and eccentricity = 4/3

Answer 1

Answer:

(x - 4)² /9 - (y - 3)² /7 = 0

Step-by-step explanation:

The equation of hyperbola = (x - h)² /a² - (y - k)² /b²  = 0

Where (h,k) is the center of the hyperbola.  

‘a’ is distance of vertex from center.  

‘e’ is eccentricity  

‘b’ is calculated as b² + a² = a²e²

Given, focii of the hyperbola are (8, 3) and (0, 3) and the eccentricity 4/3

Now, the center of the hyperbola is the mid-point of the line joining two foci i

So, Coordinate of center (h, k) = {(8 + 0)/2, (3 + 3)/2} = (4, 3)

Now, the distance between the focii = √{(0 - 8)² + (3 - 3)²} = √(8)² = 8

2a * e = 8  (eccentricity = e =4/3)

a * (4/3) = 4  

a = 3

a² = 9

Now, b² = a²(e² - 1)

b² = 9{(4/3)² - 1}

= 9(16/9 - 1)

= 9 * (7/9)

=> b² = 7

Equation of hyperbola is given by (x - h)² /a² - (y - k)² /b² = 0

(x - 4)² /9 - (y - 3)² /7 = 0

Answer 2

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=7(x-4)^{2}-9(y-3)^{2}=63}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (8,3)\:and\:(0,3)} \\ \\ \tt{ : \implies Eccentricity(e) = \frac{4}{3}} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2ae } \\ \\ \tt{ : \implies Foci \ = \sqrt{(8-0)^{2}+(3-3)^{2}}=8} \\ \\ \tt{ : \implies 2ae = 8} \\ \\ \tt{: \implies a \times \frac{4}{3} = 4} \\ \\ \green{ \tt{: \implies a = 3} }\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {b}^{2} = {3}^{2} ( {(\frac{4}{3})}^{2} - 1)} \\ \\ \tt{ : \implies {b}^{2} = 9\times \frac{7}{9}} \\ \\ \green{\tt{: \implies {b}^{2} = 7}} \\ \\ \tt{:\implies Co-ordinate\:of\:centre=\frac{8+0}{2},\frac{3+3}{2}}\\\\ \tt{:\implies Co-ordinate\:of\:centre=(4,3)}\\ \\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies \frac{ {(x-4)}^{2} }{9} - \frac{ {(y-3)}^{2} }{7} = 1}} \\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: 7{(x-4)}^{2} - 9({y-3)}^{2} = 63}}$