If the tangent to the parabola y^2 = ax makes | vedclass

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(D) The given parabola is $y^2 = ax$,which can be written as $y^2 = 4 \left( \frac{a}{4} \right) x$. Comparing this with the standard form $y^2 = 4Ax$

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$\left( \frac{a}{4}, \frac{a}{2} \right)$

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(D) The given parabola is $y^2 = ax$,which can be written as $y^2 = 4 \left( \frac{a}{4} ight) x$. Comparing this with the standard form $y^2 = 4Ax$,we have $A = \frac{a}{4}$. The slope of the tangent $m$ is given by $	an(45^{\circ}) = 1$. For a parabola $y^2 = 4Ax$,the slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{2A}{y_1}$. Substituting the values,$1 = \frac{2(a/4)}{y_1} = \frac{a/2}{y_1}$. Thus,$y_1 = \frac{a}{2}$. Since the point $(x_1, y_1)$ lies on the parabola $y^2 = ax$,we have $y_1^2 = ax_1$. Substituting $y_1 = \frac{a}{2}$,we get $\left( \frac{a}{2} ight)^2 = ax_1$,which implies $\frac{a^2}{4} = ax_1$. Therefore,$x_1 = \frac{a}{4}$. The point of contact is $\left( \frac{a}{4}, \frac{a}{2} ight)$.

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

If the tangent to the parabola $y^2 = ax$ makes an angle of $45^{\circ}$ with the $x$-axis,then the point of contact is

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