The locus of a point z in the Argand plane th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $F_1$ be the point $(1, -1)$ and $F_2$ be the point $(2, 1)$.The given equation is $|z - F_1| - |z - F_2| = 3$,which represents the difference

Vedclass · Accepted Answer

none of the above

Vedclass · Answer

(D) Let $F_1$ be the point $(1, -1)$ and $F_2$ be the point $(2, 1)$.The given equation is $|z - F_1| - |z - F_2| = 3$,which represents the difference of distances from two fixed points.The distance between the foci $F_1$ and $F_2$ is $d = |F_1F_2| = \sqrt{(2 - 1)^2 + (1 - (-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$.For a hyperbola defined by $|PF_1 - PF_2| = 2a$,the condition $2a Here,$2a = 3$ and $d = \sqrt{5} \approx 2.236$.Since $3 > \sqrt{5}$,the condition $2a Therefore,no such point $z$ exists in the Argand plane that satisfies the given equation.

The locus of a point $z$ in the Argand plane that moves satisfying the equation $|z - (1 - i)| - |z - (2 + i)| = 3$ is:

Similar Questions

Let $a, b \in \mathbb{R}$ and the roots $\alpha, \beta$ of the equation $z^2+az+b=0$ be complex. If the origin,$\alpha$ and $\beta$ represent the vertices of an equilateral triangle on the Argand plane,then

If $z_1, z_2, z_3$ are vertices of a triangle in the Argand plane such that $|z_1 - z_2| = |z_1 - z_3|$,then $\arg \left( \frac{2z_1 - z_2 - z_3}{z_3 - z_2} \right)$ is:

$Z$ is a complex number such that $|Z| \leq 2$ and $-\frac{\pi}{3} \leq \operatorname{amp} Z \leq \frac{\pi}{3}$. The area of the region formed by the locus of $Z$ is (in sq. units)

If $|z - 2 - 3i| + |z + 2 - 6i| = 4$,where $i = \sqrt{-1}$,then the locus of $P(z)$ is

Let $z$ be a complex number such that $\left|\frac{z-2i}{z+i}\right|=2$,where $z \neq -i$. Then $z$ lies on a circle of radius $2$ and center:

Vedclass Test Series

Exam Paper Generator

Online Exam Module