Let the normal at the point P on the parabola | vedclass

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(B) The equation of the parabola is $y^{2} = 6x$,so $4a = 6$,which gives $a = \frac{3}{2}$.The equation of the normal at point $P(at^{2}, 2at)$ is $y

Vedclass · Accepted Answer

$-\frac{9}{4}$

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(B) The equation of the parabola is $y^{2} = 6x$,so $4a = 6$,which gives $a = \frac{3}{2}$.The equation of the normal at point $P(at^{2}, 2at)$ is $y = -tx + 2at + at^{3}$.Since the normal passes through $(5, -8)$,we have $-8 = -t(5) + 2(\frac{3}{2})t + \frac{3}{2}t^{3}$.$-8 = -5t + 3t + \frac{3}{2}t^{3} \implies -8 = -2t + \frac{3}{2}t^{3} \implies 3t^{3} - 4t + 16 = 0$.Testing for roots,$t = -2$ satisfies the equation: $3(-8) - 4(-2) + 16 = -24 + 8 + 16 = 0$.So,$t = -2$. The point $P$ is $(a(-2)^{2}, 2a(-2)) = (4a, -4a) = (4 	imes \frac{3}{2}, -4 	imes \frac{3}{2}) = (6, -6)$.The equation of the tangent at $P(6, -6)$ to $y^{2} = 6x$ is $yy_{1} = 2a(x + x_{1})$.$y(-6) = 3(x + 6) \implies -6y = 3x + 18 \implies x + 2y + 6 = 0$.The directrix of the parabola $y^{2} = 6x$ is $x = -a = -\frac{3}{2}$.To find the point $Q$,substitute $x = -\frac{3}{2}$ into the tangent equation:$-\frac{3}{2} + 2y + 6 = 0 \implies 2y = \frac{3}{2} - 6 = -\frac{9}{2} \implies y = -\frac{9}{4}$.Thus,the ordinate of $Q$ is $-\frac{9}{4}$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

Let the normal at the point $P$ on the parabola $y^{2} = 6x$ pass through the point $(5, -8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$,then the ordinate of the point $Q$ is

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