Question

Let $Z_1,Z_2,Z_3$ be three points $A,B$ and $P$ respectively in the argand plane. Let $P$ moves in the plane such that $|Z-Z_1|+|Z-Z_2|=K.$ Let $Z_1=–Z_2=i$ Maximum area of the triangle ABP is (if K = 4)

• A. 2 sq. units
• B. $\sqrt{3}$ sq. units
• C. $2\sqrt{3}$ sq. units
• D. 4 sq. units

Answer 1

The correct option is B

$\sqrt{3}$ sq. units

$|Z-Z_1|+|Z-Z_2|=K$ is the condition of an ellipse.

It is a standing ellipse with foci $(0,1)$ and $(0,-1)$

$\Rightarrow be=1$ [Comparing it with standard equation of ellipse : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity, $e$]

$\Rightarrow b\times \sqrt{1-\frac{a^2}{b^2}}=1$

$\Rightarrow \sqrt{b^2-a^2}=1$

$\Rightarrow b^2-a^2=1$

Now, $b=2\because K=4$

$\Rightarrow a=\sqrt{3}$

$AB=2$

Area of the triangle $ABP$ is maximum when $P$ is at $(\sqrt{3},0)$ or $(-\sqrt{3},0)$

So, area of the triangle $ABP=\frac{1}{2}\times AB\times \sqrt{3}=\sqrt{3}$ sq. units