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Question Number 182514 by Best1 last updated on 11/Dec/22
1. prove that the equaion of parabola    whose axis of symmetry is parallel   to y axisis given as (x−h)^2 =+_−  4p(y−k)  2.   if Dis a perpendicular distance      of the point p(r,t)from the line      (L):(x/r)+(y/t)=1  then find the         value of D   why not helped me??
$$\mathrm{1}.\:{prove}\:{that}\:{the}\:{equaion}\:{of}\:{parabola} \\ $$$$\:\:{whose}\:{axis}\:{of}\:{symmetry}\:{is}\:{parallel}\: \\ $$$${to}\:{y}\:{axisis}\:{given}\:{as}\:\left({x}−{h}\right)^{\mathrm{2}} =\underset{−} {+}\:\mathrm{4}{p}\left({y}−{k}\right) \\ $$$$\mathrm{2}.\:\:\:{if}\:{Dis}\:{a}\:{perpendicular}\:{distance}\: \\ $$$$\:\:\:{of}\:{the}\:{point}\:{p}\left({r},{t}\right){from}\:{the}\:{line} \\ $$$$\:\:\:\:\left({L}\right):\frac{{x}}{{r}}+\frac{{y}}{{t}}=\mathrm{1}\:\:{then}\:{find}\:{the} \\ $$$$\:\:\:\:\:\:\:{value}\:{of}\:{D} \\ $$$$\:{why}\:{not}\:{helped}\:{me}?? \\ $$
Commented by mr W last updated on 11/Dec/22
do you mean people must help you,  no matter if they can or can′t and if  they will or won′t? and they should  tell why if they don′t (can′t or won′t)  help you?
$${do}\:{you}\:{mean}\:{people}\:{must}\:{help}\:{you}, \\ $$$${no}\:{matter}\:{if}\:{they}\:{can}\:{or}\:{can}'{t}\:{and}\:{if} \\ $$$${they}\:{will}\:{or}\:{won}'{t}?\:{and}\:{they}\:{should} \\ $$$${tell}\:{why}\:{if}\:{they}\:{don}'{t}\:\left({can}'{t}\:{or}\:{won}'{t}\right) \\ $$$${help}\:{you}? \\ $$
Answered by mr W last updated on 11/Dec/22
2)  distance from point (u,v) to line  ax+bx+c=0 is d=((∣au+bv+c∣)/( (√(a^2 +b^2 )))).  applying this you can get  D=((∣(r/r)+(t/t)−1∣)/( (√((1/r^2 )+(1/t^2 )))))=((∣rt∣)/( (√(r^2 +t^2 ))))
$$\left.\mathrm{2}\right) \\ $$$${distance}\:{from}\:{point}\:\left({u},{v}\right)\:{to}\:{line} \\ $$$${ax}+{bx}+{c}=\mathrm{0}\:{is}\:{d}=\frac{\mid{au}+{bv}+{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}. \\ $$$${applying}\:{this}\:{you}\:{can}\:{get} \\ $$$${D}=\frac{\mid\frac{{r}}{{r}}+\frac{{t}}{{t}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\frac{\mid{rt}\mid}{\:\sqrt{{r}^{\mathrm{2}} +{t}^{\mathrm{2}} }} \\ $$

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