Question Number 213276 by thetpainghtun_111 last updated on 02/Nov/24
$$\mathrm{y}^{\mathrm{2}} \:=\:−\:\mathrm{4px} \\ $$$$\:\mathrm{At}\:\left(−\frac{\mathrm{1}}{\mathrm{3}},\mathrm{1}\right)\rightarrow\:\mathrm{1}=\:−\mathrm{4p}\:\left(−\:\frac{\mathrm{1}}{\mathrm{3}}\:−\:\mathrm{h}\right) \\ $$$$\:\mathrm{At}\:\left(−\frac{\mathrm{5}}{\mathrm{3}},\mathrm{2}\right)\rightarrow\:\mathrm{4}\:=\:−\mathrm{4p}\:\left(−\:\frac{\mathrm{5}}{\mathrm{3}}\:−\:\mathrm{h}\right) \\ $$$$\:\:\:\:\:\mathrm{4}\:=\:\frac{−\:\frac{\mathrm{5}}{\mathrm{3}}\:−\:\mathrm{h}}{−\:\frac{\mathrm{1}}{\mathrm{3}}\:−\:\mathrm{h}} \\ $$$$\:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\:+\:\mathrm{4h}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\:+\:\mathrm{h}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3h}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:=\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$
Commented by TonyCWX08 last updated on 02/Nov/24
$${What}\:{is}\:{this}? \\ $$
Commented by MrGaster last updated on 02/Nov/24
My personal understanding is that this is a parabola problem.
Answered by MrGaster last updated on 02/Nov/24
$$\mathrm{1}=−\mathrm{4}{p}\left(−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$\mathrm{1}=−\mathrm{4p}\left(−\frac{\mathrm{3}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$\mathrm{1}=−\mathrm{4p}\left(−\frac{\mathrm{4}}{{p}}\right) \\ $$$$\mathrm{1}=\mathrm{4}{p}\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{9}=\mathrm{16}{p} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{{p}=\frac{\mathrm{9}}{\mathrm{16}}}\\\hline\end{array} \\ $$