The straight line x + y = sqrt{2}p will touch | vedclass

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(D) The given hyperbola is $4x^2 - 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The line is

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$2p^2 = 5$

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(D) The given hyperbola is $4x^2 - 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The line is $x + y = \sqrt{2}p$,which can be written as $y = -x + \sqrt{2}p$. Comparing this with $y = mx + c$,we have $m = -1$ and $c = \sqrt{2}p$. The condition for the line $y = mx + c$ to touch the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$. Substituting the values,we get $(\sqrt{2}p)^2 = 9(-1)^2 - 4$. $2p^2 = 9 - 4$. $2p^2 = 5$.

The straight line $x + y = \sqrt{2}p$ will touch the hyperbola $4x^2 - 9y^2 = 36$,if

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