Question

An equilateral triangle is inscribed in the parabola $y^2=4ax$ with one vertex at the origin. The radius of the circum circle of that triangle is $ka$ then find $k$ .

Answer 1

Answer: 8

Let, $△OAB$ is the given triangle of side length $l$, then $AB$ is the double ordinate of the parabola.
Also, $OA$ makes an angle of $30^{\circ }$, with the $X$ - axis, hence the coordinate of the point $A$ are $\left(\ell \cos 30^{\circ },\ell \sin 30^{\circ }\right)=\left(\frac{\ell \sqrt{3}}{2},\frac{\ell }{2}\right)$ And,this point lies on the parabola, then it must satisfy the equation of the parabola $y^2=4ax$
As $\Rightarrow {\left(\frac{\ell }{2}\right)}^2=4a\left(\ell \frac{\sqrt{3}}{2}\right)$
$\Rightarrow \frac{\ell }{4}=\frac{4a\sqrt{3}}{2}$
$\Rightarrow \ell =8a\sqrt{3}...\left(i\right)$
We know that, for an equilateral triangle the circumcentre, incentre, centroid and orthocentre are same, hence the radius of the circumcircle $ka$ is $\frac{2}{3}$ of the length of the altitude of the triangle.
Also, length of the altitude of the triangle with side length $\ell$ is $\frac{\sqrt{3}}{2}\ell$
Hence, $ka=\frac{2}{3}\times \left(\frac{\sqrt{3}}{2}\right)\ell$
$\Rightarrow ka=\left(\frac{1}{\sqrt{3}}\right)\ell$
From equation $\left(i\right)$ put the value of $\ell ,$ to get
$ka=\frac{1}{\sqrt{3}}\times \left(8a\sqrt{3}\right)$
$\Rightarrow k=8$