The lengths of the two focal chords of the pa | vedclass

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(D) For a parabola $y^2 = 4ax$,the length of a focal chord making an angle $	heta$ with the axis is $L = 4a \csc^2 	heta$.Given $4a = 16$,so $a = 4$

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(D) For a parabola $y^2 = 4ax$,the length of a focal chord making an angle $	heta$ with the axis is $L = 4a \csc^2 	heta$.Given $4a = 16$,so $a = 4$. The length $L = 25$.Thus,$25 = 16 \csc^2 	heta \implies \csc^2 	heta = \frac{25}{16} \implies \sin^2 	heta = \frac{16}{25} \implies \sin 	heta = \pm \frac{4}{5}$.Let the two chords be at angles $	heta$ and $\pi - 	heta$ (or $-	heta$) with the axis. The area of the quadrilateral formed by the endpoints of two focal chords is given by the formula $Area = \frac{1}{2} |(x_1 - x_2)(y_1 + y_2)|$ or more simply $Area = 2a^2 \sin(2	heta) \csc^4 	heta$ is not standard,but the area of a quadrilateral formed by two focal chords is $Area = \frac{1}{2} L_1 L_2 \sin \phi$,where $\phi$ is the angle between the chords.Here $L_1 = L_2 = 25$. The angle between the chords is $2	heta$. Since $\sin 	heta = \frac{4}{5}$,$\cos 	heta = \frac{3}{5}$.Then $\sin(2	heta) = 2 \sin 	heta \cos 	heta = 2 	imes \frac{4}{5} 	imes \frac{3}{5} = \frac{24}{25}$.Area $= \frac{1}{2} 	imes 25 	imes 25 	imes \frac{24}{25} = \frac{1}{2} 	imes 25 	imes 24 = 300$ sq. units.

The lengths of the two focal chords of the parabola $y^2 = 16x$ are $25$ units each. If these two chords cut the parabola at $A, B, C$ and $D$,then the area (in sq. units) of the quadrilateral formed by $A, B, C$ and $D$ is

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