Which of the following lines is a normal to t | vedclass

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(D) For the parabola $y^2 = 4ax$,we have $a = 4$.The equation of a normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$

Vedclass · Accepted Answer

$y = (x - 11) \cos 	heta - \cos 3	heta$

Vedclass · Answer

(D) For the parabola $y^2 = 4ax$,we have $a = 4$.The equation of a normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.Alternatively,the equation of a normal in terms of slope $m$ is $y = mx - 2am - am^3$.Given the options,we test the form $y = m(x - c_1) - c_2$. Comparing with the options,let $m = \cos 	heta$.The equation of the normal is $y = m(x - 2a) - am^3$.Substituting $a = 4$ and $m = \cos 	heta$:$y = \cos 	heta (x - 2(4)) - 4 \cos^3 	heta$$y = (x - 8) \cos 	heta - 4 \cos^3 	heta$Using the identity $\cos 3	heta = 4 \cos^3 	heta - 3 \cos 	heta$,we have $4 \cos^3 	heta = \cos 3	heta + 3 \cos 	heta$.Substituting this into the equation:$y = x \cos 	heta - 8 \cos 	heta - (\cos 3	heta + 3 \cos 	heta)$$y = x \cos 	heta - 11 \cos 	heta - \cos 3	heta$$y = (x - 11) \cos 	heta - \cos 3	heta$.Thus,option $D$ is correct.

Which of the following lines is a normal to the parabola $y^2 = 16x$?

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