The equation of the tangent to the curve 6y = | vedclass

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

Question

(B) The given curve is $6y = 7 - x^3$. To find the slope of the tangent, we differentiate the equation with respect to $x$: $6 \frac{dy}{dx} = -3x^2$

Vedclass · Accepted Answer

$x + 2y = 3$

Vedclass · Answer

(B) The given curve is $6y = 7 - x^3$. To find the slope of the tangent, we differentiate the equation with respect to $x$: $6 \frac{dy}{dx} = -3x^2$ $\frac{dy}{dx} = -\frac{3x^2}{6} = -\frac{x^2}{2}$ At the point $(1, 1)$, the slope $m$ is: $m = -\frac{(1)^2}{2} = -\frac{1}{2}$ The equation of the tangent line passing through $(x_1, y_1) = (1, 1)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$: $y - 1 = -\frac{1}{2}(x - 1)$ $2(y - 1) = -(x - 1)$ $2y - 2 = -x + 1$ $x + 2y = 3$

The equation of the tangent to the curve $6y = 7 - x^3$ at the point $(1, 1)$ is:

Similar Questions

Let $n \in (0, \infty)$. If all the curves $y = x^n \log x$ for distinct values of $n$ always have $y = x - 1$ as the tangent drawn at a fixed point $(\alpha, \beta)$,then $\alpha + \beta =$

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $(-2,0)$ and cuts the $Y$-axis at a point $Q$ where its gradient is $3$. Then the values of $a, b, c$ respectively are:

Find the point on the curve $y=x^{3}-11x+5$ at which the tangent is $y=x-11$.

If the line $ax + by + c = 0$ is a tangent to the curve $xy = 4$, then

If at any point on a curve the subtangent and subnormal are equal,then the length of the normal is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module