The equation of the tangent at A(2, 3) to the | vedclass

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

Question

(A) Given the curve $y = ax^3 + b$ and the point $A(2, 3)$ lies on the curve,we have: $3 = a(2)^3 + b \implies 8a + b = 3$  (Equation $1$). The slope

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(A) Given the curve $y = ax^3 + b$ and the point $A(2, 3)$ lies on the curve,we have: $3 = a(2)^3 + b \implies 8a + b = 3$  (Equation $1$). The slope of the tangent to the curve at any point $x$ is given by the derivative $\frac{dy}{dx} = 3ax^2$. At $x = 2$,the slope is $m = 3a(2)^2 = 12a$. The equation of the tangent is given as $y = 4x - 5$,which is in the form $y = mx + c$,so the slope $m = 4$. Equating the slopes: $12a = 4 \implies a = \frac{4}{12} = \frac{1}{3}$. Substitute $a = \frac{1}{3}$ into Equation $1$: $8(\frac{1}{3}) + b = 3$ $\frac{8}{3} + b = 3$ $b = 3 - \frac{8}{3} = \frac{9 - 8}{3} = \frac{1}{3}$. Thus,$b = \frac{1}{3}$.

The equation of the tangent at $A(2, 3)$ to the curve $y = ax^3 + b$ is $y = 4x - 5$. Then $b =$

Similar Questions

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 = \dots$

An angle between the curves $x^2=3y$ and $x^2+y^2=4$ is

If the lengths of the tangent,subtangent,normal and subnormal for the curve $y=x^2+x-1$ at the point $(1,1)$ are $a, b, c$ and $d$ respectively,then their increasing order is

On the curve $y=x^3$,the point at which the tangent line is parallel to the chord joining the points $(-1, -1)$ and $(2, 8)$ is

The normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis. Then $f^{\prime}(3)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module