Question

Let $PQ:2x+y+6=0$ is a chord of the curve $x^2-4y^2=4.$ Coordinates of the point $R(\alpha ,\beta )$ that satisfy ${\alpha }^2+{\beta }^2-1\le 0;$ such that area of triangle $PQR$ is minimum, are given by :

• A. $(\frac{-2}{\sqrt{5}},\frac{1}{\sqrt{5}})$
• B. $(\frac{-2}{\sqrt{5}},\frac{-1}{\sqrt{5}})$
  
• C. $(\frac{-2}{\sqrt{5}},\frac{1}{\sqrt{5}})$
• D. $(\frac{2}{\sqrt{5}},\frac{-1}{\sqrt{5}})$

Answer 1

The correct option is B $(\frac{-2}{\sqrt{5}},\frac{-1}{\sqrt{5}})$


$\frac{x^2}{4}-\frac{y^2}{1}=1$
$R(\alpha ,\beta )$ lies on or inside circle $x^2+y^2=1$

$PQ$ length is fixed. So,triangle of minimum area will be formed only when the perpendicular distance from point $R$ is least.
Any tangent to the given circle is
$x\cos \theta +y\sin \theta =1$
$\because$ it has to be parallel to $PQ$
$\therefore \cot \theta =2$
So, point of contact will be
$(\cos \theta ,\sin \theta )\equiv (\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}),(-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}})$
Hence for minimum distance $R\equiv (\frac{-2}{\sqrt{5}},\frac{-1}{\sqrt{5}})$