Question Number 175176 by nadovic last updated on 21/Aug/22
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\: \\ $$$$\mathrm{points}\:\mathrm{equidistant}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$${A}\left(\mathrm{4},\:−\mathrm{1}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:{x}−{y}+\mathrm{2}=\mathrm{0}. \\ $$
Answered by som(math1967) last updated on 22/Aug/22
![let movale point(h,k) from the condition (√((h−4)^2 +(k+1)^2 ))=((∣h−k+2∣)/( (√((1)^2 +(−1)^2 )))) 2(h−4)^2 +2(k+1)^2 =h^2 +k^2 +4−2hk +4h−4k [squaring both side] h^2 +k^2 +12h−8k−2hk−30=0 ∴ equation of the locus x^2 +y^2 +12x−8y−2xy−30=0 locus path is parabola](https://www.tinkutara.com/question/Q175178.png)
$${let}\:{movale}\:{point}\left({h},{k}\right) \\ $$$${from}\:{the}\:{condition} \\ $$$$\sqrt{\left({h}−\mathrm{4}\right)^{\mathrm{2}} +\left({k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mid{h}−{k}+\mathrm{2}\mid}{\:\sqrt{\left(\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{2}\left({h}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} +\mathrm{4}−\mathrm{2}{hk} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{h}−\mathrm{4}{k} \\ $$$$\:\:\left[{squaring}\:{both}\:{side}\right] \\ $$$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} +\mathrm{12}{h}−\mathrm{8}{k}−\mathrm{2}{hk}−\mathrm{30}=\mathrm{0} \\ $$$$\therefore\:{equation}\:{of}\:{the}\:{locus} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\mathrm{12}\boldsymbol{{x}}−\mathrm{8}\boldsymbol{{y}}−\mathrm{2}\boldsymbol{{xy}}−\mathrm{30}=\mathrm{0} \\ $$$$\boldsymbol{{locus}}\:\boldsymbol{{path}}\:\boldsymbol{{is}}\:\boldsymbol{{parabola}} \\ $$
Commented by nadovic last updated on 22/Aug/22
$${Thank}\:{you}\:{Sir} \\ $$
Answered by a.lgnaoui last updated on 22/Aug/22
$$ \\ $$$${l}\:{ensemble}\:{des}\:{points}\:{situes}\:{a}\:{la}\:{meme}\:{distance}\:{de}\:{A}\:{a}\:{la}\:{droite}\:\left({D}\right) \\ $$$${y}={x}+\mathrm{2}\:{au}\:{pont}\:{M}\left(\mathrm{4},−\mathrm{1}\right)\:{sont}\:{les}\:{points}\:{de}\:{la}\:{droite}\:{parallele}\:{a}\:\left({D}\right){note}\:\left({D}'\right)\:{qui}\:{s} \\ $$$${qui}\:{passe}\:{par}\:{M}\:\:\:\:{y}={x}+{m} \\ $$$${x}=\mathrm{4}\:\:\:\:\:\:{y}=−\mathrm{1} \\ $$$$\mathrm{4}+{m}=−\mathrm{1}\:\:\:\:\:{m}=−\mathrm{5} \\ $$$${y}={x}−\mathrm{5}\:\:\: \\ $$
Commented by som(math1967) last updated on 22/Aug/22
$$\left(\mathrm{4},−\mathrm{1}\right)\:{not}\:{equidistace}\:{to}\:{y}={x}+\mathrm{2} \\ $$$${and}\:{y}={x}−\mathrm{5} \\ $$