The line x cos alpha + y sin alpha = p will t | vedclass

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Question

(A) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be written as $x \cos \alpha + y \sin \alpha - p = 0$. The equation of t

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$p \cos \alpha + a = 0$

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(A) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be written as $x \cos \alpha + y \sin \alpha - p = 0$. The equation of the parabola is $y^2 = 4a(x + a)$. The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4ax'$ (where $x' = x+a$) is $c = \frac{a'}{m}$. Rewriting the line equation: $y \sin \alpha = -x \cos \alpha + p \implies y = -x \cot \alpha + p \csc \alpha$. Here,$m = -\cot \alpha$ and $c = p \csc \alpha$. The parabola is $y^2 = 4a(x+a)$. Comparing with $Y^2 = 4aX$,we have $Y = y$,$X = x+a$,and the parameter $a$ remains $a$. The condition for tangency is $c = \frac{a}{m} + am$ (since the vertex is at $(-a, 0)$). Substituting the values: $p \csc \alpha = \frac{a}{-\cot \alpha} + a(-\cot \alpha)$. $p \csc \alpha = -a 	an \alpha - a \cot \alpha = -a (\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}) = -a (\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}) = -\frac{a}{\sin \alpha \cos \alpha}$. $p \frac{1}{\sin \alpha} = -\frac{a}{\sin \alpha \cos \alpha} \implies p = -\frac{a}{\cos \alpha}$. Thus,$p \cos \alpha = -a$,or $p \cos \alpha + a = 0$.

The line $x \cos \alpha + y \sin \alpha = p$ will touch the parabola $y^2 = 4a(x + a)$ if:

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