If y = 2x + 3 is a tangent to the parabola y^ | vedclass

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(C) The equation of the parabola is $y^2 = 24x$,so $4a = 24$,which gives $a = 6$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ i

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$15\sqrt{5}$

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(C) The equation of the parabola is $y^2 = 24x$,so $4a = 24$,which gives $a = 6$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.Since the normal is parallel to the tangent $y = 2x + 3$,its slope $m$ must be $2$.Substituting $a = 6$ and $m = 2$ into the normal equation:$y = 2x - 2(6)(2) - 6(2)^3$$y = 2x - 24 - 48$$y = 2x - 72$.The distance between two parallel lines $y = mx + c_1$ and $y = mx + c_2$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}$.Here,$c_1 = 3$,$c_2 = -72$,and $m = 2$.$d = \frac{|3 - (-72)|}{\sqrt{1 + 2^2}} = \frac{75}{\sqrt{5}} = \frac{75\sqrt{5}}{5} = 15\sqrt{5}$.

If $y = 2x + 3$ is a tangent to the parabola $y^2 = 24x$,what is the distance between the tangent and the parallel normal?

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