Find the length of the chord of the parabola | vedclass

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(B) The equation of the parabola is $y^2 = 4ax$,where $a = 1$.The chord passes through the vertex $(0, 0)$ and makes an angle $	heta = 45^{\circ}$ wi

Vedclass · Accepted Answer

$4\sqrt{2}$

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(B) The equation of the parabola is $y^2 = 4ax$,where $a = 1$.The chord passes through the vertex $(0, 0)$ and makes an angle $	heta = 45^{\circ}$ with the $x$-axis.The slope of the chord is $m = 	an(45^{\circ}) = 1$.The equation of the line passing through $(0, 0)$ with slope $m = 1$ is $y = x$.Substitute $y = x$ into the parabola equation $y^2 = 4x$:$x^2 = 4x$$x^2 - 4x = 0$$x(x - 4) = 0$So,$x = 0$ or $x = 4$.If $x = 0$,then $y = 0$. The point is $(0, 0)$.If $x = 4$,then $y = 4$. The point is $(4, 4)$.The length of the chord is the distance between $(0, 0)$ and $(4, 4)$:$L = \sqrt{(4 - 0)^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.

Find the length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes an angle of $45^{\circ}$ with the $x$-axis.

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