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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Ellipse

Vedclass · Answer

(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

Similar Questions

The eccentricity of the conic $36x^2 + 144y^2 - 36x - 96y - 119 = 0$ is

Let $P$ be an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a > b > 0$. Suppose $F_1$ and $F_2$ are the foci of the ellipse. The locus of the centroid of the $\triangle P F_1 F_2$ as $P$ moves on the ellipse is

The equation of the ellipse with eccentricity $e = \frac{1}{2}$ and foci at $(\pm 1, 0)$ is

The foci of the ellipse $2x^2 + 3y^2 - 4x - 12y + 13 = 0$ are

If the tangent to the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is parallel to the $x$-axis,then what is the point of tangency?

Vedclass Test Series

Exam Paper Generator

Online Exam Module