The ratio of the distance of a point from a f | vedclass

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(B) Let the point be $P(h, k)$. The fixed point (focus) is $S(-2, 0)$ and the directrix is $x = 9/2$,which can be written as $2x - 9 = 0$.Given the ra

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(B) Let the point be $P(h, k)$. The fixed point (focus) is $S(-2, 0)$ and the directrix is $x = 9/2$,which can be written as $2x - 9 = 0$.Given the ratio of distances $PS : PM = 2 : 3$,where $PM$ is the perpendicular distance from $P$ to the line $2x - 9 = 0$.So,$PS = \frac{2}{3} PM$.Squaring both sides,$(PS)^2 = \frac{4}{9} (PM)^2$.$(h + 2)^2 + k^2 = \frac{4}{9} \left( \frac{2h - 9}{2} \right)^2$.$(h + 2)^2 + k^2 = \frac{4}{9} \cdot \frac{(2h - 9)^2}{4}$.$9[(h + 2)^2 + k^2] = (2h - 9)^2$.$9[h^2 + 4h + 4 + k^2] = 4h^2 - 36h + 81$.$9h^2 + 36h + 36 + 9k^2 = 4h^2 - 36h + 81$.$5h^2 + 9k^2 = 45$.Dividing by $45$,we get $\frac{h^2}{9} + \frac{k^2}{5} = 1$.The locus of the point $P(h, k)$ is $\frac{x^2}{9} + \frac{y^2}{5} = 1$,which represents an ellipse since the eccentricity $e = 2/3 < 1$.

The ratio of the distance of a point from a fixed point and a line $x = 9/2$ is always $2 : 3$. Then the locus of the point is

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