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(C) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $\Delta = abc + 2fgh - a

Vedclass · Accepted Answer

$57$

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(C) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.For the given equation $2x^2 - xy + ky^2 + 6x + y + 4 = 0$,we have $a=2, h=-1/2, b=k, g=3, f=1/2, c=4$.The condition for a pair of lines is $\Delta = 2(4k - 1/4) + 1/2(-2 - 3/2) + 3(-1/4 - 3k) = 0$.Simplifying,$8k - 1/2 - 7/4 - 3/4 - 9k = 0$,which gives $-k - 3 = 0$,so $k = -3$.The pair of lines is $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$,which factors as $(2x - 3y + 4)(x + y + 1) = 0$.The perpendicular distances from $(-1, 5)$ to these lines are $d_1 = \frac{|2(-1) - 3(5) + 4|}{\sqrt{2^2 + (-3)^2}} = \frac{|-13|}{\sqrt{13}} = \sqrt{13}$ and $d_2 = \frac{|-1 + 5 + 1|}{\sqrt{1^2 + 1^2}} = \frac{5}{\sqrt{2}}$.The product is $\sqrt{13} 	imes \frac{5}{\sqrt{2}} = \frac{5\sqrt{26}}{2} = \frac{65}{\sqrt{26}}$,which matches the given condition.Finally,$37k^2 + 92k = 37(-3)^2 + 92(-3) = 37(9) - 276 = 333 - 276 = 57$.

If the product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy + ky^2 + 6x + y + 4 = 0$ is $\frac{65}{\sqrt{26}}$,then $37k^2 + 92k$ is equal to

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