Tangent and normal are drawn at P(16, 16) on | vedclass

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(A) The equation of the parabola is ${y^2} = 16x$,so $4a = 16$,which implies $a = 4$.For the point $P(16, 16)$,we have $y_1^2 = 16x_1$,so $16^2 = 16(1

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$2$

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(A) The equation of the parabola is ${y^2} = 16x$,so $4a = 16$,which implies $a = 4$.For the point $P(16, 16)$,we have $y_1^2 = 16x_1$,so $16^2 = 16(16)$,which is satisfied.The tangent at $P(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$,so $16y = 8(x + 16)$,which simplifies to $2y = x + 16$.Setting $y = 0$ for the intersection with the $x$-axis,we get $x = -16$,so $A = (-16, 0)$.The normal at $P(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{2a}(x - x_1)$,so $y - 16 = -\frac{16}{8}(x - 16)$,which simplifies to $y - 16 = -2(x - 16)$,or $y = -2x + 48$.Setting $y = 0$ for the intersection with the $x$-axis,we get $2x = 48$,so $x = 24$,thus $B = (24, 0)$.The circle passes through $A(-16, 0)$,$B(24, 0)$,and $P(16, 16)$. Since $A$ and $B$ lie on the $x$-axis,the center $C$ must have an $x$-coordinate of $\frac{-16 + 24}{2} = 4$.Let $C = (4, k)$. Since $CA = CP$,we have $(4 - (-16))^2 + (k - 0)^2 = (4 - 16)^2 + (k - 16)^2$.$20^2 + k^2 = (-12)^2 + (k - 16)^2 \Rightarrow 400 + k^2 = 144 + k^2 - 32k + 256$.$400 = 400 - 32k \Rightarrow k = 0$. Thus,$C = (4, 0)$.Slope of $PC$ $(m_1)$ = $\frac{16 - 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$.Slope of $PB$ $(m_2)$ = $\frac{16 - 0}{16 - 24} = \frac{16}{-8} = -2$.$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{\frac{4}{3} - (-2)}{1 + (\frac{4}{3})(-2)} ight| = \left| \frac{\frac{10}{3}}{1 - \frac{8}{3}} ight| = \left| \frac{\frac{10}{3}}{-\frac{5}{3}} ight| = |-2| = 2$.

Tangent and normal are drawn at $P(16, 16)$ on the parabola ${y^2} = 16x$,which intersect the axis of the parabola at $A$ and $B$,respectively. If $C$ is the centre of the circle through the points $P, A$ and $B$ and $\angle CPB = \theta$,then a value of $\tan \theta$ is:

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