The tangent to the curve $y = ax^2 + bx$ at $(2, -8)$ is parallel to the $x$-axis. Then:

  • A
    $a = 2, b = -2$
  • B
    $a = 2, b = -4$
  • C
    $a = 2, b = -8$
  • D
    $a = 4, b = -4$

Explore More

Similar Questions

The sum of the intercepts on the axes cut off by the tangent to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$ is:

If the tangent at point $(1, 2)$ on the curve $y = ax^2 + bx + \frac{7}{2}$ is parallel to the normal at $(-2, 2)$ on the curve $y = x^2 + 6x + 10$,then:

Let $f: R \rightarrow R$ be a bijection. $A$ curve represented by $y=f(x)$ is such that $f^{\prime}(x)>0$ for all $x \in R$. The tangent and normal drawn at $P(\alpha, 1)$ on the curve cut the $X$-axis at $A$ and $B$ respectively,and $C$ is the foot of the perpendicular from $P$ onto the $X$-axis. If $P(\alpha, 1)$ is such a point that $AC+CB$ is minimum,then the tangent at $P$ is parallel to the line

The equation of the tangent to the curve $y=4 x e^{x}$ at $\left(-1, -\frac{4}{e}\right)$ is

If the normal to the curve $y = x + \frac{2}{x}$ at the point where the abscissa is $2$ meets the coordinate axes at points $A$ and $B$,find the length of $AB$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo