Question

Let $\Gamma$ be a circle with diameter $AB$ and centre $O$. Let $l$ be the tangent to $\Gamma$ at $B$. For each point $M$ on $\Gamma$ different from $A$, consider the tangent $t$ at $M$ and let it intersect $l$ at $P$. Draw a line parallel to $AB$ through $P$ intersecting $OM$ at $Q$. The locus of $Q$ as $M$ varies over $\Gamma$ is

• A. an arc of a circle
• B. a parabola
• C. an arc of an ellipse
• D. a branch of a hyperbola

Answer 1

The correct option is B a parabola

Let assume the equation of circle be : $x^2+y^2=a^2$


Point $M=(a\cos \theta ,a\sin \theta )$
Tangent at $M:x\cos \theta +y\sin \theta =a$.....(i)
Tangent at $B:x=a$.........(ii)
Finding point P, using equation (i) and (ii)
$y\sin \theta =a-a\cos \theta \Rightarrow y=\frac{a(1-\cos \theta )}{\sin \theta }\Rightarrow y=a\tan \frac{\theta }{2}$
$\therefore P=(a,a\tan \frac{\theta }{2})$
Equation of line $OM:y=\tan \theta .x$...(iii)
Equation of line $PQ:y=a\tan \frac{\theta }{2}$....(iv)
Let the point $Q=(h,k)$
Finding point $Q$, using equation (iii) and (iv)
$k=a\tan \frac{\theta }{2}\text{and}h=\frac{a\tan \frac{\theta }{2}}{\tan \theta }$
Finding the locus of point $Q$,
$\tan \theta =\frac{k}{h}\Rightarrow \frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\frac{k}{h}\Rightarrow 2\frac{\frac{k}{a}}{1-(\frac{k}{a}{)}^2}=\frac{k}{h}\Rightarrow \frac{2a}{a^2-k^2}=\frac{1}{h}\Rightarrow 2ax=a^2-y^2$
Hence the locus of Q will be a parabola