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