The equation of the straight line which passe | vedclass

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(B) Given point $(x_{1}, y_{1}) = (a \cos^{3} 	heta, a \sin^{3} 	heta)$.Given line: $x \sec 	heta + y \operatorname{cosec} 	heta = a$.Rewriting th

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$x \cos 	heta - y \sin 	heta = a \cos 2 	heta$

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(B) Given point $(x_{1}, y_{1}) = (a \cos^{3} 	heta, a \sin^{3} 	heta)$.Given line: $x \sec 	heta + y \operatorname{cosec} 	heta = a$.Rewriting the given line in slope-intercept form:$y \operatorname{cosec} 	heta = -x \sec 	heta + a$$y = -\frac{\sec 	heta}{\operatorname{cosec} 	heta} x + \frac{a}{\operatorname{cosec} 	heta}$$y = -\frac{\sin 	heta}{\cos 	heta} x + a \sin 	heta$.Thus,the slope $m_{1} = -\frac{\sin 	heta}{\cos 	heta}$.The slope $m$ of the line perpendicular to this line is $m = -\frac{1}{m_{1}} = \frac{\cos 	heta}{\sin 	heta}$.The equation of the required line is $(y - y_{1}) = m(x - x_{1})$:$y - a \sin^{3} 	heta = \frac{\cos 	heta}{\sin 	heta} (x - a \cos^{3} 	heta)$$y \sin 	heta - a \sin^{4} 	heta = x \cos 	heta - a \cos^{4} 	heta$$x \cos 	heta - y \sin 	heta = a \cos^{4} 	heta - a \sin^{4} 	heta$$x \cos 	heta - y \sin 	heta = a (\cos^{2} 	heta - \sin^{2} 	heta) (\cos^{2} 	heta + \sin^{2} 	heta)$Since $\cos^{2} 	heta + \sin^{2} 	heta = 1$ and $\cos^{2} 	heta - \sin^{2} 	heta = \cos 2 	heta$,we get:$x \cos 	heta - y \sin 	heta = a \cos 2 	heta$.

The equation of the straight line which passes through the point $(a \cos^{3} \theta, a \sin^{3} \theta)$ and is perpendicular to $x \sec \theta + y \operatorname{cosec} \theta = a$ is

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