Question

find locus of point p for which the distance of point p to (6,5) is triple the distance from p to x-axis

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{The distance of P from (6,5) is triple the distance from P to the x axis}$

$\underline{\textsf{To find:}}$

$\textsf{The locus of P}$

$\underline{\textsf{Solution:}}$

$\textsf{Let the co-ordinates of P be (h,k) and A be the point (6,5)}$

$\mathsf{As\;per\;given\;data,}$

$\mathsf{AP=3{\times}k}$

$\implies\mathsf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=3k}$

$\implies\mathsf{\sqrt{(h-6)^2+(k-5)^2}=3k}$

$\mathsf{Squaring\;on\;bothsides\;,we get}$

$\implies\mathsf{(h-6)^2+(k-5)^2=9k^2}$

$\implies\mathsf{h^2+36-12h+k^2+25-10k=9k^2}$

$\implies\mathsf{h^2+36-12h-8k^2+25-10k=0}$

$\implies\mathsf{h^2-8k^2-12h-10k+61=0}$

$\therefore\mathsf{The;\;locus\;of\;P\;is\;\;\boxed{\mathsf{x^2-8y^2-12x-10y+61=0}}}$

Find more:

B and c are fixed points having coordinates (3,0) and (-3,0) respectively. If the vertical angle bac is 90 then locus

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Answer 2

$Let P be (x,y)</p><p></p><p>Distance from (4,0)</p><p></p><p>(x−4)2+y2x2−4x+16+y2x2+y2−4x+16⋯(1)</p><p></p><p>Distance from Origin</p><p></p><p>x2+y2</p><p></p><p>According to Question</p><p></p><p></p><p>x2+y2−4x+16=2x2+y2</p><p></p><p>Squaring on both sides</p><p></p><p>(x2+y2−4x+16)=4(x2+y2)3x2+3y2+4x−16=0</p><p></p><p>$