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(C) The cosine of the angle $	heta$ between the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\cos 	heta = \frac{|a+b|}{\sqrt{(

Vedclass · Accepted Answer

$A-III, B-I, C-V, D-IV$

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(C) The cosine of the angle $	heta$ between the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\cos 	heta = \frac{|a+b|}{\sqrt{(a-b)^2 + 4h^2}}$.$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$: Here $a=5, h=\sqrt{7}, b=-1$. $\cos 	heta = \frac{|5-1|}{\sqrt{(5-(-1))^2 + 4(\sqrt{7})^2}} = \frac{4}{\sqrt{36+28}} = \frac{4}{\sqrt{64}} = \frac{4}{8} = \frac{1}{2}$. Thus,$A-III$.$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$: Here $a=1, h=\frac{\sqrt{11}}{2}, b=2$. $\cos 	heta = \frac{|1+2|}{\sqrt{(1-2)^2 + 4(\frac{\sqrt{11}}{2})^2}} = \frac{3}{\sqrt{1+11}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$. Thus,$B-I$.$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$: Here $a=1, h=\sqrt{2}, b=1$. $\cos 	heta = \frac{|1+1|}{\sqrt{(1-1)^2 + 4(\sqrt{2})^2}} = \frac{2}{\sqrt{0+8}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$. Thus,$C-V$.$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$: Here $a=3, h=2\sqrt{2}, b=1$. $\cos 	heta = \frac{|3+1|}{\sqrt{(3-1)^2 + 4(2\sqrt{2})^2}} = \frac{4}{\sqrt{4+32}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3}$. Thus,$D-IV$.Therefore,the correct match is $A-III, B-I, C-V, D-IV$.

List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

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