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(A) Given the pair of lines $a^2 x^2 + 2hxy + b^2 y^2 = 0$. Let the slopes of the lines be $m$ and $2m$. From the properties of quadratic equations in

Vedclass · Accepted Answer

$\frac{3 \sqrt{2}}{4}$

Vedclass · Answer

(A) Given the pair of lines $a^2 x^2 + 2hxy + b^2 y^2 = 0$. Let the slopes of the lines be $m$ and $2m$. From the properties of quadratic equations in $m$ (where $y = mx$),we have: Sum of slopes: $m + 2m = -\frac{2h}{b^2}$ $\Rightarrow 3m = -\frac{2h}{b^2}$ $\Rightarrow m = -\frac{2h}{3b^2} \dots (i)$ Product of slopes: $m 	imes 2m = \frac{a^2}{b^2} \Rightarrow 2m^2 = \frac{a^2}{b^2} \dots (ii)$ Substituting $(i)$ into $(ii)$: $2 \left(-\frac{2h}{3b^2}ight)^2 = \frac{a^2}{b^2}$ $2 \left(\frac{4h^2}{9b^4}ight) = \frac{a^2}{b^2}$ $\frac{8h^2}{9b^4} = \frac{a^2}{b^2}$ $\frac{h^2}{a^2 b^2} = \frac{9}{8}$ Taking the square root (since $a, b, h > 0$): $\frac{h}{ab} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4}$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

For $a, b, h > 0$,if the slope of one of the lines represented by $a^2 x^2 + 2hxy + b^2 y^2 = 0$ is twice that of the other,then the value of $\frac{h}{ab}$ is

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