Question

The length of the transverse axis of a hyperbola is 7 and it passes through the point (5. - 2). The equation of the hyperbola -

Answer 1

Answer:

4/49 x^2 - 51/196 y^2 = 1

Step-by-step explanation:

Given The length of the transverse axis of a hyperbola is 7 and it passes through the point (5. - 2). The equation of the hyperbola

Let the equation of the hyperbola be x^2/a^2 - y^2/b^2 = 1

We know that length of transverse axis is 2a = 7 and so a = 7/2

The point (5, - 2) lies on hyperbola

So we have x = 5 and a = 7/2 and y = - 2. We need to find b

So 25/(7/2)^2 - 4/b^2 = 1

100/49 - 4/b^2 = 1

b^2 = 49 x 4 /51

b^2 = 196/51

Now we have the equation of parabola that is

4/49 x^2 - 51/196 y^2 = 1

Answer 2

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=16x^{2}-51y^{2}=196}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Transverse\: axis = 7} \\ \\ \tt{:\implies Point\:of\:hyperbola=(5,-2)} \\ \\ \red {\underline \bold{To \: Find: }} \\ \tt {: \implies Eqn \: of \:hyperbola=?}$

• According to given question :

$\bold{As \: we \: know \: that} \\\tt{:\implies Transverse\: axis = 2a} \\ \\ \tt{: \implies 7= 2a} \\ \\ \green{\tt{: \implies a = \frac{7}{2} }} \\ \\ \bold{Eqn \: of \: standard \: hyperbola} \\ \tt{: \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1 } \\ \\ \text{Putting \: value \: of \: b} \\ \tt{: \implies \frac{ {x}^{2} }{ (\frac{7}{2})^{2} } - \frac{ {y}^{2} }{ b^{2} } = 1} \\ \\ \tt{: \implies \frac{ 4{x}^{2} }{ 49} - \frac{ {y}^{2} }{b^{2}} = 1} \\ \\ \text{(5, - 2) \: satisfy \: on \: this \: eqn} \\ \tt{: \implies \frac{ 4\times{5}^{2} }{49 } - \frac{ {( - 2)}^{2} }{b^{2}} = 1} \\ \\ \tt{: \implies \frac{100}{49 } - \frac{4}{b^{2}} = 1} \\ \\ \tt{: \implies \frac{4}{ {b}^{2} } = \frac{100}{49 } -1}\\ \\ \tt{: \implies \frac{4}{ {b}^{2} } = \frac{51}{49} } \\ \\ \tt{: \implies 51 {b}^{2} = 196} \\ \\ \tt{: \implies {b}^{2} = \frac{196}{51} }$

$\text{Putting \: value \: of \: a \: and \: b \: in \: standard \: eqn} \\ \tt{: \implies \frac{ 4{x}^{2} }{49} - \frac{51 {y}^{2} }{196} = 1} \\ \\ \green{ \tt{ : \implies 16 {x}^{2} - 51 {y}^{2} = 196}} \\ \\ \green{\tt{\therefore Eqn \: of \: hyperbola \: is \: \: 16 {x}^{2} - 51 {y}^{2} = 196}}$