Quote from

Hvmaths on September 2, 2020, 10:42 pm

We know that locus of a point which is equidistant from fixed point is equal to its distance from fixed line is a parabola.

With AB as directrix and centroid O as focus.

Equation of parabola is

$\cfrac{a}{\sqrt{3}}y=-x^2+\cfrac{a^2}{12}$

And equation of OA is $\sqrt{3}y=x$

Hence coordinator of point Q are

$\left( \cfrac{a}{6},\cfrac{a}{6\sqrt{3}}\right) $

Hence Required Area is

=6(Area of Region OPQ)

$=6\int_{0}^{\frac{a}{6}} \cfrac{\sqrt{3}}{a} (\cfrac{a^2}{12} - x^2) dx $

$=\cfrac{2\sqrt{3}}{27}a^2 $

We know that locus of a point which is equidistant from fixed point is equal to its distance from fixed line is a parabola.

With AB as directrix and centroid O as focus.

Equation of parabola is

$\cfrac{a}{\sqrt{3}}y=-x^2+\cfrac{a^2}{12}$

And equation of OA is $\sqrt{3}y=x$

Hence coordinator of point Q are

$\left( \cfrac{a}{6},\cfrac{a}{6\sqrt{3}}\right) $

Hence Required Area is

=6(Area of Region OPQ)

$=6\int_{0}^{\frac{a}{6}} \cfrac{\sqrt{3}}{a} (\cfrac{a^2}{12} - x^2) dx $

$=\cfrac{2\sqrt{3}}{27}a^2 $

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