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Question-190019




Question Number 190019 by ajfour last updated on 26/Mar/23
Commented by ajfour last updated on 26/Mar/23
Given p and q. Find r.   Parabola is y=x^2 .
$${Given}\:{p}\:{and}\:{q}.\:{Find}\:{r}.\: \\ $$$${Parabola}\:{is}\:{y}={x}^{\mathrm{2}} . \\ $$
Commented by mr W last updated on 26/Mar/23
you have given  r=((√(q^2 −p^2 ))/2)
$${you}\:{have}\:{given} \\ $$$${r}=\frac{\sqrt{{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 26/Mar/23
oops!, forget q being givwn.
$${oops}!,\:{forget}\:{q}\:{being}\:{givwn}. \\ $$
Answered by ajfour last updated on 26/Mar/23
A(−s,s^2 )    ; tan α=2s  B(t,t^2 )  ; tan β=2t  −s+rsin α=pcos θ  s^2 +rcos α=psin θ  −pcos θ+2rsin θ+rsin β=t  psin θ+2rcos θ−rcos β=t^2   ..
$${A}\left(−{s},{s}^{\mathrm{2}} \right)\:\:\:\:;\:\mathrm{tan}\:\alpha=\mathrm{2}{s} \\ $$$${B}\left({t},{t}^{\mathrm{2}} \right)\:\:;\:\mathrm{tan}\:\beta=\mathrm{2}{t} \\ $$$$−{s}+{r}\mathrm{sin}\:\alpha={p}\mathrm{cos}\:\theta \\ $$$${s}^{\mathrm{2}} +{r}\mathrm{cos}\:\alpha={p}\mathrm{sin}\:\theta \\ $$$$−{p}\mathrm{cos}\:\theta+\mathrm{2}{r}\mathrm{sin}\:\theta+{r}\mathrm{sin}\:\beta={t} \\ $$$${p}\mathrm{sin}\:\theta+\mathrm{2}{r}\mathrm{cos}\:\theta−{r}\mathrm{cos}\:\beta={t}^{\mathrm{2}} \\ $$$$.. \\ $$

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