Question

A point moves so that its distance from the point $(2,0)$ is always $\frac{1}{3}$ of its distance from the line $x-18=0.$ If the locus of the point is a conic, then

• A. Locus of the point will be $\frac{x^2}{36}+\frac{y^2}{32}=1.$
• B. Locus of the point will be $\frac{x^2}{18}+\frac{y^2}{16}=1.$
• C. Length of the latus rectum of the conic $=\frac{32}{3}$ units
• D. Length of the latus rectum of the conic $=\frac{16\sqrt{2}}{3}$ units

Answer 1

Answer: (A), (C)

The correct options are
A Locus of the point will be $\frac{x^2}{36}+\frac{y^2}{32}=1.$
C Length of the latus rectum of the conic $=\frac{32}{3}$ units

From the basic definition of the conic $(2,0)$ will be focus and $x=18$ will be directrix and $e=\frac{1}{3}$ will be eccentricity
and here $e<1$ hence conic is ellipse
$\because$ Focus is on $x-$axis and directrix is $\perp x-$axis
$\therefore$ It is a standard horizontal ellipse, so
$ae=2,\frac{a}{e}=18\Rightarrow a^2=36\therefore b^2=a^2(1-e^2)=32$

Hence, the equation of ellipse is
$\frac{x^2}{36}+\frac{y^2}{32}=1$

Length of latus rectum
$=\frac{2b^2}{a}=\frac{2\times 32}{6}=\frac{32}{3}\text{units}$