Playing with ellipse

Geometry Level 2

An ellipse $x^2$ +2 $y^2$ = $a^2$ has eccentricity 'e'.

Let 'P' be a point inside the ellipse such that it is at a distance 6√2 and 8√2 from the ends of the minor axis.

'P' is at a distance ae from the origin .

Find the value of $\frac{a}{5}$

This is an original problem and belongs to my set Raju bhai's creations

1 solution

Rajath Rao
Nov 12, 2017

The given data is represented in the above diagram.

Standard equation of ellipse $\implies \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ………1

The given ellipse $\implies \dfrac{x^2}{a^2}+\dfrac{y^2}{\frac{a^2}{2}}=1$ ………2

On comparing 1 and 2.

$b^2=\dfrac{a^2}{2}$ ………3

$2b^2=a^2$

$b^2=a^2-b^2$

$b^2=\dfrac{a^2(a^2-b^2)}{a^2}$

$b=ae$ (Because , $e=\sqrt{\dfrac{a^2-b^2}{a^2}}$ )

Therefore, PO = AO = BO

We can construct a circle with A,P,F,B,F' as its points.

So, APB=90° because,(angle in a semicircle is always equal to 90°)

Given, AP=6√2 , BP=8√2

By pythagorous theorem, AB=10√2

AB=2b=10√2

b=5√2

From 3,

$(5\sqrt{2})^2=\dfrac{a^2}{2}$

On simplifying,

a=10

Therefore, $\boxed{\dfrac{a}{5}=2}$