The equation of the tangent to the curve y=1- | vedclass

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

Question

(B) Given equation of the curve is $y=1-e^{\frac{x}{3}} \dots (i)$.Since the curve intersects the $Y$-axis,we set $x=0$.Substituting $x=0$ in $(i)$,we

Vedclass · Accepted Answer

$x+3y=0$

Vedclass · Answer

(B) Given equation of the curve is $y=1-e^{\frac{x}{3}} \dots (i)$.Since the curve intersects the $Y$-axis,we set $x=0$.Substituting $x=0$ in $(i)$,we get $y=1-e^{0}=1-1=0$.Thus,the point of intersection is $(0, 0)$.The slope of the tangent is given by $\frac{dy}{dx}$.Differentiating $(i)$ with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{3} e^{\frac{x}{3}}$.At the point $(0, 0)$,the slope $m = \left(\frac{dy}{dx}\right)_{(0,0)} = -\frac{1}{3} e^{0} = -\frac{1}{3}$.The equation of the tangent line passing through $(0, 0)$ with slope $m = -\frac{1}{3}$ is $y - 0 = -\frac{1}{3}(x - 0)$.Multiplying by $3$,we get $3y = -x$,which simplifies to $x + 3y = 0$.

The equation of the tangent to the curve $y=1-e^{\frac{x}{3}}$ at the point of intersection with the $Y$-axis is

Similar Questions

The equation of the tangent to the curve $y^{2}=ax^{2}+b$ at the point $(2,3)$ is $y=4x-5$. Then the values of $a$ and $b$ are:

If the equation of a tangent drawn to the curve $y=\cos (x+y), -1 \leq x \leq 1+\pi$ is $x+2y=k$, then $k=$

The area of the triangle formed by the normal to the curve $y = e^{2x} + x^2$ at the point $(0, 1)$ with the coordinate axes is $......$ square units.

Let $f(x) = \begin{cases} -x^2 & \text{for } x < 0 \\ x^2 + 8 & \text{for } x \ge 0 \end{cases}$. Then the $x$-intercept of the line that is tangent to the graph of $f(x)$ is

The length of the subnormal at any point on the curve $y = (\frac{x}{2024})^k$ is constant if the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module