For what value of gamma does the line y = 2x | vedclass

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(A) The equation of the hyperbola is $16x^{2} - 9y^{2} = 144$. Dividing by $144$,we get $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$. Here,$a^{2} = 9$ and

Vedclass · Accepted Answer

$\pm 2\sqrt{5}$

Vedclass · Answer

(A) The equation of the hyperbola is $16x^{2} - 9y^{2} = 144$. Dividing by $144$,we get $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$. Here,$a^{2} = 9$ and $b^{2} = 16$. The line $y = mx + c$ touches the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ if $c^{2} = a^{2}m^{2} - b^{2}$. Comparing $y = 2x + \gamma$ with $y = mx + c$,we have $m = 2$ and $c = \gamma$. Substituting the values into the condition: $\gamma^{2} = (9)(2^{2}) - 16$. $\gamma^{2} = (9)(4) - 16 = 36 - 16 = 20$. Therefore,$\gamma = \pm \sqrt{20} = \pm 2\sqrt{5}$.

For what value of $\gamma$ does the line $y = 2x + \gamma$ touch the hyperbola $16x^{2} - 9y^{2} = 144$?

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