The equation of the tangent to the parabola y | vedclass

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(D) The equation of the tangent to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx+\frac{a}{m}$.Here,the parabola is $y^2=8x$,so $4a=8$,which

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$x-\sqrt{3}y+6=0$

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(D) The equation of the tangent to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx+\frac{a}{m}$.Here,the parabola is $y^2=8x$,so $4a=8$,which implies $a=2$.The inclination is $30^{\circ}$,so the slope $m=	an(30^{\circ})=\frac{1}{\sqrt{3}}$.Substituting these values into the tangent equation:$y=\frac{1}{\sqrt{3}}x+\frac{2}{1/\sqrt{3}}$$y=\frac{x}{\sqrt{3}}+2\sqrt{3}$Multiplying by $\sqrt{3}$ on both sides:$\sqrt{3}y=x+6$$x-\sqrt{3}y+6=0$.Thus,the correct option is $D$.

The equation of the tangent to the parabola $y^2=8x$ inclined at $30^{\circ}$ to the $X$-axis is

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