If varphi is the angle between the lines a x^ | vedclass

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Question

(A) The angle $\varphi$ between the lines represented by $a x^{2}+2 h x y+b y^{2}=0$ is given by $	an \varphi = \left| \frac{2 \sqrt{h^{2}-a b}}{a+b}

Vedclass · Accepted Answer

$	heta$

Vedclass · Answer

(A) The angle $\varphi$ between the lines represented by $a x^{2}+2 h x y+b y^{2}=0$ is given by $	an \varphi = \left| \frac{2 \sqrt{h^{2}-a b}}{a+b} ight|$.For the equation $x^{2}+2 x y \sec 	heta+y^{2}=0$,we have $a=1$,$b=1$,and $h=\sec 	heta$.Substituting these values into the formula:$	an \varphi = \left| \frac{2 \sqrt{\sec^{2} 	heta - (1)(1)}}{1+1} ight|$$	an \varphi = \left| \frac{2 \sqrt{	an^{2} 	heta}}{2} ight|$$	an \varphi = 	an 	heta$.Therefore,the angle between the lines is $	heta$.

If $\varphi$ is the angle between the lines $a x^{2}+2 h x y+b y^{2}=0$,then the angle between the lines $x^{2}+2 x y \sec \theta+y^{2}=0$ is

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