Let x=9 be a directrix of an ellipse E , whose centre is at the origin and eccentricity is…

Question

Let $x=9$ be a directrix of an ellipse E , whose centre is at the origin and eccentricity is $\frac{1}{3}$. Let $\mathrm{P}(\alpha, 0)$, $\alpha>0$, be a focus of E and AB be a chord passing through P . Then the locus of the mid point of AB...

Accepted Answer

Since the centre of the ellipse is at the origin and one directrix is $x=9$, the major axis is along the $x$-axis.

Let the ellipse be

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$

with eccentricity

$e=\frac{1}{3}$

For an ellipse with major axis along the $x$-axis, the directrices are

$x=\pm \frac{a}{e}$

Given one directrix is $x=9$, so

$\frac{a}{e}=9$

Since $e=\frac{1}{3}$,

$a=9\cdot \frac{1}{3}=3$

Also,

$c=ae=3\cdot \frac{1}{3}=1$

So the focus $P(\alpha ,0)$ is

$P=(1,0)$

Hence $\alpha =1$.

Now,

${b}^{2}={a}^{2}(1-{e}^{2})=9(1-\frac{1}{9})=9\cdot \frac{8}{9}=8$

Therefore the ellipse is

$\frac{{x}^{2}}{9}+\frac{{y}^{2}}{8}=1$

Let the midpoint of chord $AB$ passing through the focus $(1,0)$ be $(h,k)$.

For the ellipse

$\frac{{x}^{2}}{9}+\frac{{y}^{2}}{8}=1$

the equation of the chord whose midpoint is $(h,k)$ is given by the NCERT midpoint formula:

$\frac{x{x}_{1}}{9}+\frac{y{y}_{1}}{8}=\frac{{x}_{2}^{1}}{9}+\frac{{y}_{2}^{1}}{8}$

So here the chord with midpoint $(h,k)$ is

$\frac{xh}{9}+\frac{yk}{8}=\frac{{h}^{2}}{9}+\frac{{k}^{2}}{8}$

Since this chord passes through the focus $(1,0)$, substitute $(x,y)=(1,0)$:

$\frac{h}{9}=\frac{{h}^{2}}{9}+\frac{{k}^{2}}{8}$

Multiply by $72$:

$8h=8{h}^{2}+9{k}^{2}$

So,

$9{k}^{2}=8h-8{h}^{2}=8h(1-h)$

Replacing $(h,k)$ by $(x,y)$ for the locus of the midpoint, we get

$9{y}^{2}=8x(1-x)$

Hence the required locus is

$9{y}^{2}=8x(1-x)$

So the correct option is A.

Solutions · 1