For the real parameter $t$ , the locus of the complex number $z = (1 - t^{2}) + i \sqrt{1 + t^{2}}$

Updated On: Jan 8, 2024

Hide Solution

Verified By Collegedunia

The Correct Option is B

Explanation:
Given

z = (1 - t^{2}) + i \sqrt{1 + t^{2}}

Let

z = x +

\begin{array}{l} x = 1 - t^{2} \\ y^{2} = 1 + t^{2} \end{array}

Thus,

x + y^{2} = 2

y^{2} = 2 - x

y^{2} = - (x - 2)

Thus parabola

Was this answer helpful?

If \[\frac{dx}{dy} = \frac{1 + x - y^2}{y}, \quad x(1) = 1,\]then $5x(2)$ is equal to:
- JEE Main - 2024
- Mathematics
- Differential equations
View Solution
Let $\alpha$ be a non-zero real number. Suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function such that $f(0) = 2$ and \[\lim_{x \to \infty} f(x) = 1.\]If $f'(x) = \alpha f(x) + 3$, for all $x \in \mathbb{R}$, then $f(-\log 2)$ is equal to ________ .
- JEE Main - 2024
- Mathematics
- Differential equations
View Solution
Let$f(x) = |2x^2 + 5|x - 3|, x \in \mathbb{R}$. If $m$ and $n$ denote the number of points were $f$is not continuous and not differentiable respectively, then$m + n$is equal to:
- JEE Main - 2024
- Mathematics
- Differential equations
View Solution
Let $ y = y(x) $ be the solution of the differential equation \[\left( 1 + x^2 \right) \frac{dy}{dx} + y = e^{\tan^{-1}x}, \, y(1) = 0.\]Then $ y(0) $ is:
- JEE Main - 2024
- Mathematics
- Differential equations
View Solution
Let $ y = y(x) $ be the solution of the differential equation \[\left( 2x \log_e x \right) \frac{dy}{dx} + 2y = \frac{3}{x} \log_e x, \, x>0 \, \text{and} \, y(e^{-1}) = 0.\] Then, $ y(e) $ is equal to:
- JEE Main - 2024
- Mathematics
- Differential equations
View Solution

View More Questions

View More Questions

COLLEGE NOTIFICATIONS

EXAM NOTIFICATIONS

NEWS UPDATES

Download the Collegedunia app on