Question

Let $P,Q,R$ be the points on the auxiliary circle of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ , such that $PQR$ is an equilateral triangle and $P^{\prime }{Q}^{\prime }{R}^{\prime }$ is corresponding triangle inscribed within the ellipse. Then centroid of the triangle $P^{\prime }{Q}^{\prime }{R}^{\prime }$ lies at

• A. centre of the ellipse
• B. focus of the ellipse
• C. between focus and centre of the ellipse
• D. between one extremity of minor axis and centre of the ellipse

Answer 1

The correct option is A centre of the ellipse

Let points on equilateral triangle be $P\left(\theta \right),Q(\theta +\frac{2\pi }{3}),R(\theta +\frac{4\pi }{3})$
then
$P^{\prime }\equiv (a\cos \theta ,b\sin \theta )Q^{\prime }\equiv (a\cos (\theta +\frac{2\pi }{3}),b\sin (\theta +\frac{2\pi }{3}))R^{\prime }\equiv (a\cos (\theta +\frac{4\pi }{3}),b\sin (\theta +\frac{4\pi }{3}))$

Let centroid of $△P^{\prime }{Q}^{\prime }{R}^{\prime }\equiv (x^{\prime },y^{\prime })$
$\therefore x^{\prime }=a\left[\frac{\cos \left(\theta \right)+\cos (\theta +\frac{2\pi }{3})+\cos (\theta +\frac{4\pi }{3})}{3}\right]\Rightarrow x^{\prime }=\frac{a}{3}[\cos \theta +2\cos (\theta +\pi \left)\cos \frac{\pi }{3}\right]=0$

And
$y^{\prime }=b\left[\frac{\sin \left(\theta \right)+\sin (\theta +\frac{2\pi }{3})+\sin (\theta +\frac{4\pi }{3})}{3}\right]\Rightarrow y^{\prime }=\frac{b}{3}[\sin \theta +2\sin (\theta +\pi \left)\cos \frac{\pi }{3}\right]=0$

Hence, centroid is $(0,0),$ which is the centre of the ellipse.