Question

MATCHI THE FOLLOWING
A point 'z' moves on the $|z-5i|+|z-11i|=10$ then match the following column.
Column I Column II
1. Maximum $|z|$ (p) $3$
2. Minimum $|z|$ (q) $\frac{\sqrt{39}}{4}$
3.$\tan (min\arg z)$ (r) $13$
4.$\tan (max\arg z)$ (s) $10+\frac{\sqrt{39}}{4}$

• A. $1\to r$
• B. $2\to p$
• C. $3\to q$
• D. $4\to q$

Answer 1

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

The correct options are
A $3\to q$
B $2\to p$
C $1\to r$
D $4\to q$

$|z-5i|+|z-11i|=10$$\Rightarrow S^{\prime }=(0,11),S=(0,5),C=(0,8)$
Here $2a=10\Rightarrow a=5$
Now $2ae=S{S}^{\prime }=6$ $\Rightarrow ae=3,a=5\Rightarrow e=\frac{3}{5}$
Now $b^2=a^2(1-e^2)=16\Rightarrow b=4$
Maximum of $\left|z\right|=13$
Minimum of $\left|z\right|=3$
Equation of the ellipse is $\frac{x^2}{16}+\frac{{(y-8)}^2}{25}=1$ ...(1)
Equation of the line passing through the argin is $y=mx$ ...(2)
From (1) and (2) $\frac{x^2}{m^2\left(16\right)}+\frac{{(y-8)}^2}{25}=1$
$\Rightarrow y^2(25+16m^2)-{\left(16\right)}^2{m}^{2}y+16m^2\left(39\right)=0$
(1) and (2) ratisfed by same ${}^{\prime }{y}^{\prime }$ values
$\therefore △=0$
$\Rightarrow m=\pm \frac{\sqrt{39}}{4}$
$\therefore \left|\tan (min\left(asgz\right))\right|=\frac{\sqrt{39}}{4}$
$\therefore \left|\tan (max\left(asgz\right))\right|=\frac{\sqrt{39}}{4}$