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Question-84512




Question Number 84512 by 698148290 last updated on 13/Mar/20
Answered by jagoll last updated on 14/Mar/20
tangent equation at point  P(2a+2t , ((at^2 )/2)) to the parabola  (x−2a)^2  = 2ay   ⇒ 2t(x−2a) = ay + (((at)^2 )/2)  2tx − 4at −ay−(1/2)(at)^2 =0
$$\mathrm{tangent}\:\mathrm{equation}\:\mathrm{at}\:\mathrm{point} \\ $$$$\mathrm{P}\left(\mathrm{2a}+\mathrm{2t}\:,\:\frac{\mathrm{at}^{\mathrm{2}} }{\mathrm{2}}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\left(\mathrm{x}−\mathrm{2a}\right)^{\mathrm{2}} \:=\:\mathrm{2ay}\: \\ $$$$\Rightarrow\:\mathrm{2t}\left(\mathrm{x}−\mathrm{2a}\right)\:=\:\mathrm{ay}\:+\:\frac{\left(\mathrm{at}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{2tx}\:−\:\mathrm{4at}\:−\mathrm{ay}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{at}\right)^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by 698148290 last updated on 14/Mar/20
Thanks
$${Thanks} \\ $$

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