Find the measure of the angle in degrees subt | vedclass

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(B) Let the parabola be $y^2 = 4ax$ with vertex $A(0, 0)$.Let the double ordinate be at $x = x_1$,so the points are $P(x_1, 2\sqrt{ax_1})$ and $Q(x_1,

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$90$

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(B) Let the parabola be $y^2 = 4ax$ with vertex $A(0, 0)$.Let the double ordinate be at $x = x_1$,so the points are $P(x_1, 2\sqrt{ax_1})$ and $Q(x_1, -2\sqrt{ax_1})$.For the latus rectum (a specific double ordinate),$x_1 = a$,so $P(a, 2a)$ and $Q(a, -2a)$.The slope of $AP$ is $m_1 = \frac{2a - 0}{a - 0} = 2$.However,the question implies the angle subtended by the latus rectum at the vertex.Slope of $AP = \frac{2a}{a} = 2$ and slope of $AQ = \frac{-2a}{a} = -2$.The angle $	heta$ between $AP$ and $AQ$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{2 - (-2)}{1 + (2)(-2)}| = |\frac{4}{1 - 4}| = |\frac{4}{-3}| = \frac{4}{3}$.Wait,if the double ordinate is such that the angle is $90^\circ$,then $m_1 m_2 = -1$. Given $m_1 = \frac{y_1}{x_1}$ and $m_2 = -\frac{y_1}{x_1}$,then $-\frac{y_1^2}{x_1^2} = -1 \implies y_1^2 = x_1^2$. Since $y_1^2 = 4ax_1$,we have $x_1^2 = 4ax_1 \implies x_1 = 4a$.Thus,for the angle to be $90^\circ$,the double ordinate must be at $x = 4a$.

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

Find the measure of the angle in degrees subtended by the double ordinate of the parabola $y^2 = 4ax$ at its vertex.

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