The length of the chord of contact of the tan | vedclass

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(C) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.Here

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$\frac{3}{2}\sqrt{41}$

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(C) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.Here,the parabola is $y^2 = 8x$,so $4a = 8$,which means $a = 2$. The point is $(2, 5)$.Substituting these values,the equation of the chord of contact is $5y = 2 	imes 2(x + 2)$,which simplifies to $5y = 4x + 8$,or $x = \frac{5y - 8}{4}$.Substituting $x$ into the parabola equation $y^2 = 8x$:$y^2 = 8 \left( \frac{5y - 8}{4} ight) = 2(5y - 8) = 10y - 16$.$y^2 - 10y + 16 = 0$,which factors as $(y - 2)(y - 8) = 0$.Thus,$y = 2$ or $y = 8$.If $y = 2$,$x = \frac{5(2) - 8}{4} = \frac{2}{4} = 0.5$. If $y = 8$,$x = \frac{5(8) - 8}{4} = \frac{32}{4} = 8$.The points of intersection are $(0.5, 2)$ and $(8, 8)$.The length of the chord is $\sqrt{(8 - 0.5)^2 + (8 - 2)^2} = \sqrt{(7.5)^2 + 6^2} = \sqrt{56.25 + 36} = \sqrt{92.25} = \sqrt{\frac{369}{4}} = \frac{3}{2}\sqrt{41}$.

The length of the chord of contact of the tangents drawn from the point $(2, 5)$ to the parabola $y^2 = 8x$ is

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