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Question Number 158708 by ajfour last updated on 07/Nov/21

Commented by ajfour last updated on 07/Nov/21

$$Findequationofparabola.$$

Answered by mr W last updated on 08/Nov/21

$${(1+p)}^2+4^2={(p+3)}^2$$$$\Rightarrow p=2$$$$2h=\frac{3}{4}\times \frac{5}{2}$$$$\Rightarrow h=\frac{15}{16}$$$$-\frac{15}{16}=A\times \frac{5}{2}\times (\frac{5}{2}-5)\Rightarrow A=\frac{3}{20}$$$$\Rightarrow y=\frac{3}{20}x(x-5)$$

Commented by mr W last updated on 08/Nov/21

Commented by mr W last updated on 08/Nov/21

$$histhehightofparabola,hereat$$$$x=\frac{5}{2}.$$

Commented by ajfour last updated on 08/Nov/21

$$2h=\frac{3}{4}\times \frac{5}{2}(how?)$$$$$$

Commented by mr W last updated on 08/Nov/21

Commented by ajfour last updated on 08/Nov/21

$$sir\tan \theta =\frac{3}{4}=\frac{h+k}{\left(5/2\right)}$$$$buthow\frac{h+k}{\left(5/2\right)}=\frac{h+h}{\left(5/2\right)}??$$

Commented by mr W last updated on 08/Nov/21

$$wecangetk=h.$$$$generallywehave$$$$\tan \theta =\frac{2b}{a}$$$$prove:$$$$y={Ax}^2$$$$y^{\prime }=2Ax$$$$atx=a,y=b$$$$b={Aa}^2$$$$\tan \theta =y^{\prime }\left(a\right)=2Aa=\frac{2b}{a}$$

Commented by mr W last updated on 08/Nov/21

Commented by ajfour last updated on 08/Nov/21

$$Excellent,thankyousir.$$

Answered by ajfour last updated on 08/Nov/21

$${(p+1)}^2+4^2={(p+3)}^2$$$$\Rightarrow p=2$$$$lety=Ax(x-5)$$$$\frac{dy}{dx}{\mid }_{x=0}=-5A=-\frac{3}{4}$$$$\Rightarrow 20y=3x(x-5)$$

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