Find-the-equation-of-the-locus-of-a-moving-point-P-such-that-its-distance-from-the-point-A-4-3-and-B-5-1-is-in-the-ratio-3-1-

Question Number 167185 by pete last updated on 09/Mar/22

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{moving}\:\mathrm{point}\:\mathrm{P}, \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{its}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{A}\left(\mathrm{4},\mathrm{3}\right)\:\mathrm{and}\:\mathrm{B}\left(\mathrm{5},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{3}\::\:\mathrm{1} \\ $$

Commented by mr W last updated on 09/Mar/22

$${the}\:{locus}\:{of}\:{a}\:{moving}\:{point},\:{whose} \\ $$$${distances}\:{to}\:{two}\:{fixed}\:{points}\:{are}\:{in} \\ $$$${a}\:{constant}\:{ratio},\:{is}\:{a}\:{circle}. \\ $$

Commented by otchereabdullai@gmail.com last updated on 21/Mar/22

$$\mathrm{Nice}! \\ $$

Answered by Rasheed.Sindhi last updated on 09/Mar/22

P=(x,y) ∣AP∣:∣BP∣=3:1 ∣AP∣=3∣BP∣ (√((4−x)^2 +(3−y)^2 ))=3(√((5−x)^2 +(1−y)^2 )) x^2 −8x+16+y^2 −6y+9=9(x^2 −10x+25+y^2 −2y+1) 8x^2 +8y^2 −82x−12y+209=0

$$\mathrm{P}=\left({x},{y}\right) \\ $$$$\mid\mathrm{AP}\mid:\mid\mathrm{BP}\mid=\mathrm{3}:\mathrm{1} \\ $$$$\mid\mathrm{AP}\mid=\mathrm{3}\mid\mathrm{BP}\mid \\ $$$$\sqrt{\left(\mathrm{4}−{x}\right)^{\mathrm{2}} +\left(\mathrm{3}−{y}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +\left(\mathrm{1}−{y}\right)^{\mathrm{2}} }\: \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}+{y}^{\mathrm{2}} −\mathrm{6}{y}+\mathrm{9}=\mathrm{9}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{25}+{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right) \\ $$$$\mathrm{8}{x}^{\mathrm{2}} +\mathrm{8}{y}^{\mathrm{2}} −\mathrm{82}{x}−\mathrm{12}{y}+\mathrm{209}=\mathrm{0} \\ $$

Answered by som(math1967) last updated on 09/Mar/22

let P(h,k) ((PA)/(PB)) =(3/1) ((√((h−4)^2 +(k−3)^2 ))/( (√((h−5)^2 +(k−1)^2 )))) =(3/1) ((h^2 −8h+16+k^2 −6k+9)/(h^2 −10h+25+k^2 −2k+1))=(9/1) 9h^2 −90h+225+9k^2 −18k+9 =h^2 −8h+k^2 −6k+25 8h^2 +8k^2 −82h−12k+209=0 equation of locus 8x^2 +8y^2 −82x−12y+209=0

$${let}\:{P}\left(\boldsymbol{{h}},\boldsymbol{{k}}\right) \\ $$$$\frac{{PA}}{{PB}}\:=\frac{\mathrm{3}}{\mathrm{1}} \\ $$$$\frac{\sqrt{\left(\boldsymbol{{h}}−\mathrm{4}\right)^{\mathrm{2}} +\left(\boldsymbol{{k}}−\mathrm{3}\right)^{\mathrm{2}} }}{\:\sqrt{\left(\boldsymbol{{h}}−\mathrm{5}\right)^{\mathrm{2}} +\left(\boldsymbol{{k}}−\mathrm{1}\right)^{\mathrm{2}} }}\:=\frac{\mathrm{3}}{\mathrm{1}} \\ $$$$\:\frac{\boldsymbol{{h}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{{h}}+\mathrm{16}+\boldsymbol{{k}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{k}}+\mathrm{9}}{\boldsymbol{{h}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{h}}+\mathrm{25}+\boldsymbol{{k}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{k}}+\mathrm{1}}=\frac{\mathrm{9}}{\mathrm{1}} \\ $$$$\mathrm{9}\boldsymbol{{h}}^{\mathrm{2}} −\mathrm{90}\boldsymbol{{h}}+\mathrm{225}+\mathrm{9}\boldsymbol{{k}}^{\mathrm{2}} −\mathrm{18}\boldsymbol{{k}}+\mathrm{9} \\ $$$$\:\:\:\:\:=\boldsymbol{{h}}^{\mathrm{2}} −\mathrm{8}{h}+{k}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{k}}+\mathrm{25} \\ $$$$\mathrm{8}\boldsymbol{{h}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{k}}^{\mathrm{2}} −\mathrm{82}\boldsymbol{{h}}−\mathrm{12}\boldsymbol{{k}}+\mathrm{209}=\mathrm{0} \\ $$$${equation}\:{of}\:{locus} \\ $$$$\mathrm{8}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{82}\boldsymbol{{x}}−\mathrm{12}\boldsymbol{{y}}+\mathrm{209}=\mathrm{0} \\ $$

Commented by pete last updated on 09/Mar/22

Thank you very much sir. But sir, why is the locus of the point not a straight line but a circle? That is my problem.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{sir},\:\mathrm{why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{not}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{but}\:\mathrm{a}\:\mathrm{circle}? \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{my}\:\mathrm{problem}. \\ $$

Commented by som(math1967) last updated on 09/Mar/22

A and B is fixed pt but P is movable pt such PA:PB=3:1 so path of P is not st. line

$${A}\:{and}\:{B}\:{is}\:{fixed}\:{pt}\:{but}\:{P}\:{is}\:{movable}\:{pt} \\ $$$${such}\:{PA}:{PB}=\mathrm{3}:\mathrm{1}\:{so}\:{path}\:{of}\:{P}\:{is}\:{not} \\ $$$${st}.\:{line}\: \\ $$

Commented by MJS_new last updated on 09/Mar/22

on a straight line (= the x−axis) to given points A & B, find P with ∣AP∣:∣BP∣=3:1 this is the same problem as this: split AB with ratio 3:1 which has exactly 2 solutions (internal and external split point) P_i , P_e these are on the line itself but if you allow “split points” in 2 dimensions, how could they all form a straight line?

$$\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\left(=\:\mathrm{the}\:{x}−\mathrm{axis}\right)\:\mathrm{to}\:\mathrm{given} \\ $$$$\mathrm{points}\:{A}\:\&\:{B},\:\mathrm{find}\:{P}\:\mathrm{with} \\ $$$$\mid{AP}\mid:\mid{BP}\mid=\mathrm{3}:\mathrm{1} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{problem}\:\mathrm{as}\:\mathrm{this}: \\ $$$$\mathrm{split}\:{AB}\:\mathrm{with}\:\mathrm{ratio}\:\mathrm{3}:\mathrm{1} \\ $$$$\mathrm{which}\:\mathrm{has}\:\mathrm{exactly}\:\mathrm{2}\:\mathrm{solutions}\:\left(\mathrm{internal}\:\mathrm{and}\right. \\ $$$$\left.\mathrm{external}\:\mathrm{split}\:\mathrm{point}\right)\:{P}_{{i}} ,\:{P}_{{e}} \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{itself}\:\mathrm{but}\:\mathrm{if}\:\mathrm{you}\:\mathrm{allow} \\ $$$$“\mathrm{split}\:\mathrm{points}''\:\mathrm{in}\:\mathrm{2}\:\mathrm{dimensions},\:\mathrm{how}\:\mathrm{could} \\ $$$$\mathrm{they}\:\mathrm{all}\:\mathrm{form}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}? \\ $$

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