The equation of the normal to the curve y = l | vedclass

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

Question

(A) Given the curve $y = \log_e x$. First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at point $P(1, 0)$. $\frac{dy}{dx}

Vedclass · Accepted Answer

$x + y = 1$

Vedclass · Answer

(A) Given the curve $y = \log_e x$. First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at point $P(1, 0)$. $\frac{dy}{dx} = \frac{1}{x}$. At point $P(1, 0)$,the slope of the tangent $m_t = \left. \frac{dy}{dx} \right|_{(1, 0)} = \frac{1}{1} = 1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal line passing through $(x_1, y_1) = (1, 0)$ with slope $m_n = -1$ is: $y - y_1 = m_n(x - x_1)$ $y - 0 = -1(x - 1)$ $y = -x + 1$ $x + y = 1$.

The equation of the normal to the curve $y = \log_e x$ at the point $P(1, 0)$ is $ . . . . . . $

Similar Questions

Find the point at which the tangent to the curve $y = \sqrt{4x - 3} - 1$ has its slope $\frac{2}{3}$.

If the length of the subnormal at any point on the curve $y = a^{1-n}x^n$ is constant,then $n = \dots$

The tangent to a non-linear curve $y = f(x)$ at any point $P(x, y)$ intersects the $x$-axis and $y$-axis at $A$ and $B$ respectively. If the normal to the curve $y = f(x)$ at $P$ intersects the $y$-axis at $C$ such that $AC = BC$,and $f(2) = 3$,then the equation of the curve is:

The equation of the tangent to the curve $y=4 x e^{x}$ at $\left(-1, -\frac{4}{e}\right)$ is

The ratio between the length of the subtangent at any point other than the origin on the parabola $y^2 = 16ax$ and the abscissa of that point is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module