Question

A circle touches the parabola $y^2=2x$ at $P(\frac{1}{2},1)$ and cuts the parabola at its vertex $V.$ If the centre of the circle is $Q,$ then

• A. the radius of the circle is $\frac{5\sqrt{2}}{4}$
• B. the radius of the circle is the maximum value of $\frac{1}{4}\sin 4x+\frac{7}{4}\cos 4x.$
• C. area of $△PVQ$ is $\frac{15}{16}$
• D. slope of $PQ$ is $-2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A the radius of the circle is $\frac{5\sqrt{2}}{4}$
B the radius of the circle is the maximum value of $\frac{1}{4}\sin 4x+\frac{7}{4}\cos 4x.$
C area of $△PVQ$ is $\frac{15}{16}$



Let the equation of circle
${(x-a)}^2+(y-b{)}^2=r^2$
where $r=\sqrt{a^2+b^2}$ (As it passes through Origin)
Slope of the tangent at $(\frac{1}{2},1)$ to the parabola $y^2=2x$
$\Rightarrow y^{\prime }=1$
Also, $2(x-a)+2(y-b)y^{\prime }=0$
$(x-a)=-1(y-b)\Rightarrow x+y=a+b$
$\Rightarrow a+b=\frac{3}{2}...\left(I\right)$
Circle passes through $(\frac{1}{2},1)$
${(\frac{1}{2}-a)}^2+(1-b{)}^2=a^2+b^2\Rightarrow a+2b=\frac{5}{4}...(II)$
Solving $\left(I\right)$ and $\left(II\right)$,
$a=\frac{7}{4},b=-\frac{1}{4}$
Coordinate of centre of circle $Q(a,b)=(\frac{7}{4},-\frac{1}{4})$
Radius of circle $r=\sqrt{\frac{49}{16}+\frac{1}{16}}=\frac{5\sqrt{2}}{4}$

Area of $\Delta PVQ=$
$\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{1}{2}& 1& 1\\ \frac{7}{4}& -\frac{1}{4}& 1\end{array}\right|=\frac{15}{16}$

The maximum value of $\frac{7}{4}\sin 4x+\frac{1}{4}\cos 4x$ is $\sqrt{{\left(\frac{7}{4}\right)}^2+{\left(\frac{1}{4}\right)}^2}=\frac{5\sqrt{2}}{4}=r$