The equation 3ax^2 - 16xy - (a^2 - 10)y^2 = 0 | vedclass

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Question

(D) The given equation is $3ax^2 - 16xy - (a^2 - 10)y^2 = 0$. Comparing this with the general form $Ax^2 + 2Hxy + By^2 = 0$,we get $A = 3a$,$2H = -16$

Vedclass · Accepted Answer

two perpendicular lines if $a = -2$

Vedclass · Answer

(D) The given equation is $3ax^2 - 16xy - (a^2 - 10)y^2 = 0$. Comparing this with the general form $Ax^2 + 2Hxy + By^2 = 0$,we get $A = 3a$,$2H = -16$ (so $H = -8$),and $B = -(a^2 - 10)$. For the equation to represent a pair of parallel lines,the condition is $H^2 = AB$. Substituting the values: $(-8)^2 = 3a(-(a^2 - 10))$ $\Rightarrow 64 = -3a^3 + 30a$ $\Rightarrow 3a^3 - 30a + 64 = 0$. For the equation to represent a pair of perpendicular lines,the condition is $A + B = 0$. Substituting the values: $3a - (a^2 - 10) = 0$ $\Rightarrow 3a - a^2 + 10 = 0$ $\Rightarrow a^2 - 3a - 10 = 0$. Factoring the quadratic: $(a - 5)(a + 2) = 0$,which gives $a = 5$ or $a = -2$. Comparing these results with the given options,option $D$ is correct.

The equation $3ax^2 - 16xy - (a^2 - 10)y^2 = 0$ represents

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