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Question-158708




Question Number 158708 by ajfour last updated on 07/Nov/21
Commented by ajfour last updated on 07/Nov/21
Find equation of parabola.
$${Find}\:{equation}\:{of}\:{parabola}. \\ $$
Answered by mr W last updated on 08/Nov/21
(1+p)^2 +4^2 =(p+3)^2   ⇒p=2  2h=(3/4)×(5/2)  ⇒h=((15)/(16))  −((15)/(16))=A×(5/2)×((5/2)−5) ⇒A=(3/(20))  ⇒y=(3/(20))x(x−5)
$$\left(\mathrm{1}+{p}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\left({p}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{p}=\mathrm{2} \\ $$$$\mathrm{2}{h}=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{15}}{\mathrm{16}} \\ $$$$−\frac{\mathrm{15}}{\mathrm{16}}={A}×\frac{\mathrm{5}}{\mathrm{2}}×\left(\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{5}\right)\:\Rightarrow{A}=\frac{\mathrm{3}}{\mathrm{20}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{3}}{\mathrm{20}}{x}\left({x}−\mathrm{5}\right) \\ $$
Commented by mr W last updated on 08/Nov/21
Commented by mr W last updated on 08/Nov/21
h is the hight of parabola, here at  x=(5/2).
$${h}\:{is}\:{the}\:{hight}\:{of}\:{parabola},\:{here}\:{at} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{2}}. \\ $$
Commented by ajfour last updated on 08/Nov/21
2h=(3/4)×(5/2)      (how?)
$$\mathrm{2}{h}=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{2}}\:\:\:\:\:\:\left({how}?\right) \\ $$$$ \\ $$
Commented by mr W last updated on 08/Nov/21
Commented by ajfour last updated on 08/Nov/21
sir  tan θ=(3/4)=((h+k)/((5/2)))  but how  ((h+k)/((5/2)))=((h+h)/((5/2)))  ??
$${sir}\:\:\mathrm{tan}\:\theta=\frac{\mathrm{3}}{\mathrm{4}}=\frac{{h}+{k}}{\left(\mathrm{5}/\mathrm{2}\right)} \\ $$$${but}\:{how}\:\:\frac{{h}+{k}}{\left(\mathrm{5}/\mathrm{2}\right)}=\frac{{h}+{h}}{\left(\mathrm{5}/\mathrm{2}\right)}\:\:?? \\ $$
Commented by mr W last updated on 08/Nov/21
we can get k=h.  generally we have  tan θ=((2b)/a)  prove:  y=Ax^2   y′=2Ax  at x=a, y=b  b=Aa^2   tan θ=y′(a)=2Aa=((2b)/a)
$${we}\:{can}\:{get}\:{k}={h}. \\ $$$${generally}\:{we}\:{have} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{b}}{{a}} \\ $$$${prove}: \\ $$$${y}={Ax}^{\mathrm{2}} \\ $$$${y}'=\mathrm{2}{Ax} \\ $$$${at}\:{x}={a},\:{y}={b} \\ $$$${b}={Aa}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta={y}'\left({a}\right)=\mathrm{2}{Aa}=\frac{\mathrm{2}{b}}{{a}} \\ $$
Commented by mr W last updated on 08/Nov/21
Commented by ajfour last updated on 08/Nov/21
Excellent, thank you sir.
$${Excellent},\:{thank}\:{you}\:{sir}. \\ $$
Answered by ajfour last updated on 08/Nov/21
(p+1)^2 +4^2 =(p+3)^2   ⇒  p=2  let  y=Ax(x−5)    (dy/dx)∣_(x=0) =−5A=−(3/4)  ⇒  20y=3x(x−5)
$$\left({p}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\left({p}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{p}=\mathrm{2} \\ $$$${let}\:\:{y}={Ax}\left({x}−\mathrm{5}\right) \\ $$$$\:\:\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{0}} =−\mathrm{5}{A}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{20}{y}=\mathrm{3}{x}\left({x}−\mathrm{5}\right) \\ $$

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