If x cos alpha + y sin alpha = p is the norma | vedclass

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(D) The given equation of the line is $x + \sqrt{3} y = -4$. To find the $X$-intercept $a$,set $y = 0$: $x = -4$,so $a = -4$. To find the $Y$-intercep

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$8 \pi$

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(D) The given equation of the line is $x + \sqrt{3} y = -4$. To find the $X$-intercept $a$,set $y = 0$: $x = -4$,so $a = -4$. To find the $Y$-intercept $b$,set $x = 0$: $\sqrt{3} y = -4$,so $b = -\frac{4}{\sqrt{3}}$. The normal form is $x \cos \alpha + y \sin \alpha = p$. Comparing $x + \sqrt{3} y = -4$ with $x \cos \alpha + y \sin \alpha = p$,we first make the constant term positive: $-x - \sqrt{3} y = 4$. Dividing by $\sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$,we get $-\frac{1}{2} x - \frac{\sqrt{3}}{2} y = 2$. Thus,$\cos \alpha = -\frac{1}{2}$ and $\sin \alpha = -\frac{\sqrt{3}}{2}$,which gives $\alpha = \frac{4 \pi}{3}$ and $p = 2$. Now,calculate $\sqrt{3} \pi b p - 3 a \alpha$: $\sqrt{3} \pi \left( -\frac{4}{\sqrt{3}} \right) (2) - 3 (-4) \left( \frac{4 \pi}{3} \right) = -8 \pi + 16 \pi = 8 \pi$.

If $x \cos \alpha + y \sin \alpha = p$ is the normal form of the equation of a straight line $x + \sqrt{3} y + 4 = 0$ and $a, b$ are respectively $X$ and $Y$ intercepts of this line,then $\sqrt{3} \pi b p - 3 a \alpha = $

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