The length of the transverse axis of a hyperb | vedclass

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(C) The length of the transverse axis is $2a = 7$,so $a = \frac{7}{2}$ and $a^2 = \frac{49}{4}$.The standard equation of a hyperbola with transverse a

Vedclass · Accepted Answer

$\frac{4}{49}x^2 - \frac{51}{196}y^2 = 1$

Vedclass · Answer

(C) The length of the transverse axis is $2a = 7$,so $a = \frac{7}{2}$ and $a^2 = \frac{49}{4}$.The standard equation of a hyperbola with transverse axis along the $x$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Substituting $a^2 = \frac{49}{4}$,we get $\frac{4x^2}{49} - \frac{y^2}{b^2} = 1$.Since the hyperbola passes through $(5, -2)$,we substitute these coordinates:$\frac{4(5)^2}{49} - \frac{(-2)^2}{b^2} = 1$$\frac{100}{49} - \frac{4}{b^2} = 1$$\frac{4}{b^2} = \frac{100}{49} - 1 = \frac{51}{49}$$b^2 = \frac{4 	imes 49}{51} = \frac{196}{51}$.Substituting $b^2$ back into the equation:$\frac{4x^2}{49} - \frac{y^2}{(196/51)} = 1$$\frac{4}{49}x^2 - \frac{51}{196}y^2 = 1$.Thus,option $C$ is correct.

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

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