Write the length of the chord of the parabola $y^{2} = 4 a x$ which passes through the vertex and is inclined to the axis $\frac{π}{4}$ .

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Solution

Let A be the vertex of the parabola.Then coordinates of A are (0,0)

Suppose AP is a chord that is inclined to an angle of $\frac{π}{4}$ radian to the X axis.Let M be the point where the perpendicular from P intersects the X axis.

Let AP=l Then, $\frac{A M}{l} = c o s \frac{π}{4}$

$\Rightarrow A M = l \times \frac{1}{\sqrt{2}}$

$\Rightarrow A M = \frac{1}{\sqrt{2}} and, \frac{P M}{l} = s i n \frac{π}{4}$

$\Rightarrow P M = l \times \frac{1}{\sqrt{2}} = \frac{l}{\sqrt{2}}$

So,the coordinates of P and $(\frac{1}{\sqrt{2}}, \frac{l}{\sqrt{2}})$ since, p lies on $y^{2} = 4 a x$

$∴ {(\frac{l}{\sqrt{2}})}^{2} = 4 a \times \frac{l}{\sqrt{2}}$

$\Rightarrow \frac{l^{2}}{2} = 4 a \times \frac{l}{\sqrt{2}}$

$\Rightarrow l = \frac{8 a}{\sqrt{2}}$

$\Rightarrow l = \frac{4 \times \sqrt{2} \times \sqrt{2 a}}{\sqrt{2}}$

$= 4 \sqrt{2}$

$\Rightarrow l = 4 \sqrt{2} a$

$∴ length of chord = 4 \sqrt{2} a$

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