Question

The least value of $p$ for which the two curves $\arg \left(z\right)=\frac{\pi }{6}$ and $\left|z-2\sqrt{3}\right|=p$ intersect is

• A. $3$
• B. $\sqrt{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $3$

Solution

Let $z=x+iy$

$\frac{y}{x}=tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}$

$x=y\sqrt{3}$

$z=y\sqrt{3}+yi$

$|z-2\sqrt{3}|=p$

$\sqrt{(y\sqrt{3}-2\sqrt{3}{)}^2+y^2}=p$

$3y^2+12-12y+y^2=p^2$

$4y^2-12y+12=p^2\dots \left(1\right)$

For this quadratic

$D=\sqrt{144-12\times 16}=-ive$

Differentiating $\left(1\right)$ w.r.t $y$

$8y-12=0$

Minimum value at $y=\frac{3}{2}$

$p^2=9-12+12$

$p^2=9$

$p=3$

A is corrcet