If the tangent at a point P on a parabola mak | vedclass

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(B) Let the parabola be $y^2 = 4ax$ with focus $S(a, 0)$.Let $P(at^2, 2at)$ be a point on the parabola.The focal distance $SP = a + at^2$.The equation

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$\alpha / 2$

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(B) Let the parabola be $y^2 = 4ax$ with focus $S(a, 0)$.Let $P(at^2, 2at)$ be a point on the parabola.The focal distance $SP = a + at^2$.The equation of the tangent at $P$ is $ty = x + at^2$.This tangent meets the axis of the parabola $(y=0)$ at point $Q(-at^2, 0)$.The distance $SQ = SO + OQ = a + at^2$.Since $SP = SQ$,the triangle $	riangle SPQ$ is an isosceles triangle.Thus,$\angle SPQ = \angle SQP = \alpha$.In $	riangle SPQ$,the exterior angle at the axis is the sum of the interior opposite angles.Let $	heta$ be the angle between the tangent and the axis. Then $	heta + \alpha = \angle SPQ = \alpha$ is not correct; rather,in $	riangle SPQ$,the exterior angle at $Q$ is $\angle SPQ + \angle PSQ = 2\alpha$ is not the case here.Actually,since $	riangle SPQ$ is isosceles with $SP=SQ$,the base angles are $\alpha$. The angle between the tangent and the axis is $	heta$. In $	riangle SPQ$,the exterior angle at $Q$ is $	heta$. Thus $	heta = \angle SPQ + \angle PSQ$. Since $\angle SPQ = \alpha$ and $\angle PSQ = \angle SPQ = \alpha$,we have $	heta = \alpha / 2$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

If the tangent at a point $P$ on a parabola makes an angle $\alpha$ with its focal distance,then the angle between the tangent and the axis of the parabola is:

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