If the tangent to the parabola 4y^2 = x makes | vedclass

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(A) The given equation of the parabola is $4y^2 = x$,which can be written as $y^2 = \frac{1}{4}x$.Comparing this with $y^2 = 4ax$,we get $4a = \frac{1

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$\left( \frac{1}{48}, \frac{1}{8\sqrt{3}} \right)$

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(A) The given equation of the parabola is $4y^2 = x$,which can be written as $y^2 = \frac{1}{4}x$.Comparing this with $y^2 = 4ax$,we get $4a = \frac{1}{4}$,so $a = \frac{1}{16}$.The slope of the tangent $m = 	an(60^{\circ}) = \sqrt{3}$.The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Substituting the values,$y = \sqrt{3}x + \frac{1/16}{\sqrt{3}} = \sqrt{3}x + \frac{1}{16\sqrt{3}}$.The point of contact for a parabola $y^2 = 4ax$ with slope $m$ is given by $\left( \frac{a}{m^2}, \frac{2a}{m} ight)$.Substituting $a = \frac{1}{16}$ and $m = \sqrt{3}$:$x = \frac{1/16}{(\sqrt{3})^2} = \frac{1}{16 	imes 3} = \frac{1}{48}$.$y = \frac{2(1/16)}{\sqrt{3}} = \frac{1/8}{\sqrt{3}} = \frac{1}{8\sqrt{3}}$.Thus,the point of contact is $\left( \frac{1}{48}, \frac{1}{8\sqrt{3}} ight)$.

If the tangent to the parabola $4y^2 = x$ makes an angle of $60^{\circ}$ with the $x$-axis,then find its point of contact.

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