The locus of the point whose ratio of distanc | vedclass

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Question

(A) Let the point be $P(x, y)$. The distance from the origin $O(0, 0)$ is $\sqrt{x^2+y^2}$.The distance from $A(-2, -3)$ is $\sqrt{(x+2)^2+(y+3)^2}$.G

Vedclass · Accepted Answer

$24(x^2+y^2)-100x-150y-325=0$

Vedclass · Answer

(A) Let the point be $P(x, y)$. The distance from the origin $O(0, 0)$ is $\sqrt{x^2+y^2}$.The distance from $A(-2, -3)$ is $\sqrt{(x+2)^2+(y+3)^2}$.Given the ratio $\frac{OP}{AP} = \frac{5}{7}$,we have $\frac{\sqrt{x^2+y^2}}{\sqrt{(x+2)^2+(y+3)^2}} = \frac{5}{7}$.Squaring both sides,we get $\frac{x^2+y^2}{(x+2)^2+(y+3)^2} = \frac{25}{49}$.$49(x^2+y^2) = 25(x^2+4x+4+y^2+6y+9)$.$49(x^2+y^2) = 25(x^2+y^2+4x+6y+13)$.$49(x^2+y^2) - 25(x^2+y^2) - 100x - 150y - 325 = 0$.$24(x^2+y^2) - 100x - 150y - 325 = 0$.Thus,option $A$ is correct.

The locus of the point whose ratio of distance from the origin to its distance from $(-2, -3)$ is $5: 7$,is given by:

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