The angle subtended by the normal chord at th | vedclass

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(C) The equation of the parabola is $y^2 = 9x$. Comparing with $y^2 = 4ax$,we get $4a = 9$,so $a = \frac{9}{4}$.The focus of the parabola is $S = (a,

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(C) The equation of the parabola is $y^2 = 9x$. Comparing with $y^2 = 4ax$,we get $4a = 9$,so $a = \frac{9}{4}$.The focus of the parabola is $S = (a, 0) = (\frac{9}{4}, 0)$.The point $P$ is $(9, 9)$. Let the normal at $P$ meet the parabola again at $Q$.The slope of the tangent at $P(x_1, y_1)$ is $m = \frac{2a}{y_1} = \frac{9/2}{9} = \frac{1}{2}$.The slope of the normal at $P$ is $m' = -2$.The equation of the normal at $P(9, 9)$ is $y - 9 = -2(x - 9)$,which simplifies to $y = -2x + 27$.Substituting $y = -2x + 27$ into $y^2 = 9x$ gives $(-2x + 27)^2 = 9x$,which is $4x^2 - 108x + 729 = 9x$,or $4x^2 - 117x + 729 = 0$.Since $x = 9$ is one root,the other root $x_2$ satisfies $9 	imes x_2 = \frac{729}{4}$,so $x_2 = \frac{81}{4}$.Then $y_2 = -2(\frac{81}{4}) + 27 = -\frac{81}{2} + \frac{54}{2} = -\frac{27}{2}$.The focus is $S = (\frac{9}{4}, 0)$. The slopes of $SP$ and $SQ$ are $m_1 = \frac{9 - 0}{9 - 9/4} = \frac{9}{27/4} = \frac{4}{3}$ and $m_2 = \frac{-27/2 - 0}{81/4 - 9/4} = \frac{-27/2}{72/4} = \frac{-27/2}{18} = -\frac{3}{4}$.Since $m_1 	imes m_2 = (\frac{4}{3}) 	imes (-\frac{3}{4}) = -1$,the angle subtended at the focus is $90^{\circ}$.

The angle subtended by the normal chord at the point $(9, 9)$ on the parabola $y^2 = 9x$ at the focus of the parabola is (in $^{\circ}$)

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