The vertices of the hyperbola 7x^2 - 49y^2 = | vedclass

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(D) The given equation of the hyperbola is $7x^2 - 49y^2 = 343$. Dividing both sides by $343$,we get: $\frac{7x^2}{343} - \frac{49y^2}{343} = \frac{34

Vedclass · Accepted Answer

$(\pm 7, 0)$

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(D) The given equation of the hyperbola is $7x^2 - 49y^2 = 343$. Dividing both sides by $343$,we get: $\frac{7x^2}{343} - \frac{49y^2}{343} = \frac{343}{343}$ $\frac{x^2}{49} - \frac{y^2}{7} = 1$. Comparing this with the standard equation of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 49$,which implies $a = 7$. The vertices of the hyperbola are given by $(\pm a, 0)$. Therefore,the vertices are $(\pm 7, 0)$. Thus,option $D$ is correct.

The vertices of the hyperbola $7x^2 - 49y^2 = 343$ are

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