Co-ordinates of a point on the curve y = x lo | vedclass

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

Question

(D) Given the curve $y = x \log x$. First,find the derivative $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The slope of the ta

Vedclass · Accepted Answer

$(e^{-2}, -2e^{-2})$

Vedclass · Answer

(D) Given the curve $y = x \log x$. First,find the derivative $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The slope of the tangent at any point $(x, y)$ is $m_t = 1 + \log x$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{1 + \log x}$. The given line is $2x - 2y = 3$,which can be written as $y = x - \frac{3}{2}$. The slope of this line is $1$. Since the normal is parallel to the line,their slopes must be equal: $-\frac{1}{1 + \log x} = 1$ $-1 = 1 + \log x$ $\log x = -2$ $x = e^{-2}$. Now,substitute $x = e^{-2}$ into the curve equation: $y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$. Thus,the required point is $(e^{-2}, -2e^{-2})$.

Co-ordinates of a point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are

Similar Questions

What is the sum of the intercepts on the coordinate axes of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at the point $(x_1, y_1)$?

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $y$-axis.

Find the angle of intersection of the curves $y^{2}=x$ and $x^{2}=y$.

If the slope of the line through $(0,0)$ which is tangent to the curve $y=x^2+x+16$ is $m$,then the value of $m-4$ is

The area (in sq. units) of the triangle formed by the tangent and normal drawn to the curve $(\frac{x}{3})^n+(\frac{y}{4})^n=2$ at $(3,4)$ and the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module