The length of the chord of the parabola y^2 = | vedclass

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(A) The equation of the chord passing through the vertex $(0, 0)$ making an angle $	heta$ with the axis ($x$-axis) is $y = x 	an 	heta$.Substitutin

Vedclass · Accepted Answer

$4a \cos 	heta \csc^2 	heta$

Vedclass · Answer

(A) The equation of the chord passing through the vertex $(0, 0)$ making an angle $	heta$ with the axis ($x$-axis) is $y = x 	an 	heta$.Substituting $y = x 	an 	heta$ into the parabola equation $y^2 = 4ax$:$(x 	an 	heta)^2 = 4ax$$x^2 	an^2 	heta = 4ax$$x(x 	an^2 	heta - 4a) = 0$So,$x = 0$ or $x = \frac{4a}{	an^2 	heta}$.For $x = \frac{4a}{	an^2 	heta}$,$y = \frac{4a}{	an^2 	heta} \cdot 	an 	heta = \frac{4a}{	an 	heta}$.The intersection points are $(0, 0)$ and $(\frac{4a}{	an^2 	heta}, \frac{4a}{	an 	heta})$.The length of the chord $L = \sqrt{(\frac{4a}{	an^2 	heta} - 0)^2 + (\frac{4a}{	an 	heta} - 0)^2}$$L = \sqrt{\frac{16a^2}{	an^4 	heta} + \frac{16a^2}{	an^2 	heta}} = \frac{4a}{	an 	heta} \sqrt{\frac{1}{	an^2 	heta} + 1}$$L = \frac{4a}{	an 	heta} \sqrt{\csc^2 	heta} = \frac{4a}{	an 	heta} \cdot \csc 	heta = 4a \cdot \frac{\cos 	heta}{\sin 	heta} \cdot \frac{1}{\sin 	heta} = 4a \cos 	heta \csc^2 	heta$.

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

The length of the chord of the parabola $y^2 = 4ax$ which passes through the vertex and makes an angle $\theta$ with the axis of the parabola is:

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