The line which is parallel to the x-axis and | vedclass

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

Question

(D) Let the line be $y = c$. The curve is $y = \sqrt{x}$.At the point of intersection $P(x, y)$,we have $y = \sqrt{x}$,so $x = y^2$. Since $y = c$,the

Vedclass · Accepted Answer

$y = \frac{1}{2}$

Vedclass · Answer

(D) Let the line be $y = c$. The curve is $y = \sqrt{x}$.At the point of intersection $P(x, y)$,we have $y = \sqrt{x}$,so $x = y^2$. Since $y = c$,the point $P$ is $(c^2, c)$.The slope of the tangent to the curve $y = \sqrt{x}$ at any point is given by $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.At point $P(c^2, c)$,the slope is $m = \frac{1}{2\sqrt{c^2}} = \frac{1}{2c}$.The angle between the curve and the line $y = c$ (which is parallel to the $x$-axis) is given as $\frac{\pi}{4}$.The slope of the line $y = c$ is $0$. The angle $	heta$ between the curve and the line is given by $	an(	heta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Here,$	an\left(\frac{\pi}{4}ight) = \left| \frac{\frac{1}{2c} - 0}{1 + \frac{1}{2c} \cdot 0} ight| = \frac{1}{2c}$.Since $1 = \frac{1}{2c}$,we get $2c = 1$,which implies $c = \frac{1}{2}$.Thus,the equation of the line is $y = \frac{1}{2}$.

The line which is parallel to the $x$-axis and intersects the curve $y = \sqrt{x}$ at an angle of $\frac{\pi}{4}$ is:

Similar Questions

Let $f: ( -\infty, \infty ) \to ( -\infty, \infty )$ be defined by $f(x) = x^3 + 1$.
Statement $1$: The function $f$ has a local extremum at $x = 0$.
Statement $2$: The function $f$ is continuous and differentiable on $( -\infty, \infty )$ and $f'(0) = 0$.

If $f$ is defined by $f(x) = \begin{cases} x & \text{for } 0 \leq x < 1 \\ 2-x & \text{for } x \geq 1 \end{cases}$,then at $x=1$,$f(x)$ is

Let $f:R \to R$ be a continuously differentiable function such that $f(2) = 6$ and $f'(2) = \frac{1}{48}$. If $\int_6^{f(x)} 4t^3 \,dt = (x - 2)g(x)$,then $\lim_{x \to 2} g(x)$ is equal to

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module