If ax^2 - y^2 + 4x - y = 0 represents a pair | vedclass

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Question

(B) The given equation is $ax^2 - y^2 + 4x - y = 0$. Comparing this with the general second-degree equation $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,w

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The given equation is $ax^2 - y^2 + 4x - y = 0$. Comparing this with the general second-degree equation $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we get: $A = a, B = -1, H = 0, G = 2, F = -\frac{1}{2}, C = 0$. For the equation to represent a pair of straight lines,the condition is $ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$. Substituting the values: $(a)(-1)(0) + 2(-\frac{1}{2})(2)(0) - a(-\frac{1}{2})^2 - (-1)(2)^2 - (0)(0)^2 = 0$. $0 + 0 - \frac{a}{4} + 4 = 0$. $\frac{a}{4} = 4$. $a = 16$.

If $ax^2 - y^2 + 4x - y = 0$ represents a pair of lines,then $a = $

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