The line 21x + 5y = k touches the hyperbola 7 | vedclass

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(A) The line $Ax + By = k$ touches the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if $k^2 = a^2 A^2 - b^2 B^2$. Given the hyperbola $7x^2 - 5y^

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$116$

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(A) The line $Ax + By = k$ touches the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if $k^2 = a^2 A^2 - b^2 B^2$. Given the hyperbola $7x^2 - 5y^2 = 232$,we can rewrite it as $\frac{x^2}{(232/7)} - \frac{y^2}{(232/5)} = 1$. Here,$a^2 = \frac{232}{7}$ and $b^2 = \frac{232}{5}$. The line is $21x + 5y = k$,so $A = 21$ and $B = 5$. Substituting these values into the condition $k^2 = a^2 A^2 - b^2 B^2$: $k^2 = \left(\frac{232}{7}ight)(21)^2 - \left(\frac{232}{5}ight)(5)^2$ $k^2 = \frac{232}{7} 	imes 441 - \frac{232}{5} 	imes 25$ $k^2 = 232 	imes 63 - 232 	imes 5$ $k^2 = 232(63 - 5) = 232 	imes 58$ $k^2 = 13456$ $k = \sqrt{13456} = 116$.

The line $21x + 5y = k$ touches the hyperbola $7x^2 - 5y^2 = 232$. Then $k =$

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