The tangent to the curve y=x^3+ax-b at the po | vedclass

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

Question

(A) Given the curve $y=x^3+ax-b$. The slope of the tangent is $\frac{dy}{dx} = 3x^2+a$. The slope of the line $y-x+4=0$ is $m_1 = 1$. Since the tangen

Vedclass · Accepted Answer

$(2,-2)$

Vedclass · Answer

(A) Given the curve $y=x^3+ax-b$. The slope of the tangent is $\frac{dy}{dx} = 3x^2+a$. The slope of the line $y-x+4=0$ is $m_1 = 1$. Since the tangent is perpendicular to the line,the slope of the tangent $m_2$ must satisfy $m_1 	imes m_2 = -1$,so $m_2 = -1$. At the point $(1,-5)$,$\frac{dy}{dx} = 3(1)^2+a = 3+a$. Setting $3+a = -1$,we get $a = -4$. Since the point $(1,-5)$ lies on the curve,we substitute $x=1, y=-5, a=-4$ into the equation: $-5 = (1)^3 + (-4)(1) - b -5 = 1 - 4 - b -5 = -3 - b b = 2$. The equation of the curve is $y = x^3 - 4x - 2$. Checking the options: For $(2,-2)$: $y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2$. Since the point $(2,-2)$ satisfies the equation,it lies on the curve.

The tangent to the curve $y=x^3+ax-b$ at the point $(1,-5)$ is perpendicular to the line $y-x+4=0$. Which one of the following points lies on the curve?

Similar Questions

What is the equation of the normal to the curve $y^2 = x^3$ at the point where the $x$-coordinate is $8$?

If the curves $y^2 = 12x - 3$ and $y^2 = 12 - kx$ cut each other orthogonally,then the length of the sub-tangent at $(1, b)$ on the curve $y^2 = 12 - kx$ is

$f(x)$ is a continuous function on $\mathbb{R}$ and $y=f(x)$ is a curve. If $(\alpha, \beta)$ is a point such that $\beta=f(\alpha)$ and $p\alpha+m\beta+n=0$ $(p \neq 0, m \neq 0)$,then which one of the following is true?

The abscissae of the points,where the tangent to the curve $y=x^{3}-3x^{2}-9x+5$ is parallel to the $x$-axis,are

What is the equation of the tangent to the curve $y = 2 \cos x$ at $x = \pi / 4$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module