The equation of the tangent to the curve y = | vedclass

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

Question

(A) The given curve is $y = b e^{-x / a}$.To find the point where the curve crosses the $Y$ axis,we set $x = 0$.Substituting $x = 0$ into the equation

Vedclass · Accepted Answer

$x/a + y/b = 1$

Vedclass · Answer

(A) The given curve is $y = b e^{-x / a}$.To find the point where the curve crosses the $Y$ axis,we set $x = 0$.Substituting $x = 0$ into the equation,we get $y = b e^0 = b$.So,the point of contact is $(0, b)$.Now,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = b \cdot e^{-x / a} \cdot (-1 / a) = -\frac{b}{a} e^{-x / a}$.At the point $(0, b)$,the slope $m$ is:$m = \left. \frac{dy}{dx} \right|_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.The equation of the tangent line passing through $(x_1, y_1) = (0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$.$y - b = -\frac{b}{a}(x - 0)$.$y - b = -\frac{b}{a}x$.Dividing both sides by $b$,we get $\frac{y}{b} - 1 = -\frac{x}{a}$.Rearranging the terms,we get $\frac{x}{a} + \frac{y}{b} = 1$.

The equation of the tangent to the curve $y = b e^{-x / a}$ at the point where it crosses the $Y$ axis is

Similar Questions

The angle between the curves $y^2=2x$ and $x^2+y^2=8$ is

The angle between the curves $y=\sin x$ and $y=\cos x$ is

Find the equation of the normals to the curve $y=x^{3}+2x+6$ which are parallel to the line $x+14y+4=0$.

The area (in square units) of the triangle formed by the $X$-axis,the tangent,and the normal drawn at $(1, 1)$ to the curve $x^3 + y^3 = 2xy$ is

If a line having slope $2$ is a tangent to the curve $y=x^4-6x^3+13x^2-10x+5$ at points $P(x_1, y_1)$ and $Q(x_2, y_2)$,where $x_1, x_2 \in \mathbb{N}$,then $x_1x_2 - y_1y_2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module