Question

The triangle formed by the common tangents to the parabola $y^2=4x$ and the circle $x^2+y^2+2x=0$ is

• A. an equilateral triangle
• B. a right angled isosceles triangle
• C. an isosceles but not right angled triangle
• D. a scalene triangle

Answer 1

The correct option is A an equilateral triangle


As circle is $(x+1{)}^2+y^2=1,$ one of the common tangent is along the $y-$axis.
Let the other common tangent has slope $m.$
Then, its equation is $y=mx+\frac{1}{m}$
or, $mx-y+\frac{1}{m}=0$
Perpendicular distance from the centre $(-1,0)$ to the tangent $mx-y+\frac{1}{m}=0$ is equal to the radius.
$\Rightarrow \left|\frac{-m+\frac{1}{m}}{\sqrt{m^2+1}}\right|=1$
$\Rightarrow {(m-\frac{1}{m})}^2=m^2+1$
$\Rightarrow \frac{1}{m^2}=3$
$\Rightarrow m=\pm \frac{1}{\sqrt{3}}$
i.e., $\angle BAO=\angle OAC=\frac{\pi }{6}$
$\Rightarrow \angle BAC=\frac{\pi }{3}$ and $AB=AC$
Hence, the triangle $ABC$ is equilateral.