The tangent to the curve $y = a x^{2} + b x$ at $(2, - 8)$ is parallel to $X$ -axis. Then,

a = 2, b = - 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a = 2, b = - 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a = 2, b = - 8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a = 4, b = - 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $a = 2, b = - 8$
Given, $y = a x^{2} + b x$
On differentiating with respect to $x$ , we get
$\frac{d y}{d x} = 2 a x + b$
At $(2, - 8),$ ${(\frac{d y}{d x})}_{(2, - 8)} = 4 a + b$
Since tangent is parallel to $X$ -axis.
Thus $\frac{d y}{d x} = 0 \Rightarrow b = - 4 a$ ....(i)
Now, point $(2, - 8)$ is on the curve $y = a x^{2} + b x$
Therefore, $- 8 = 4 a + 2 b$ ....(ii)
On solving equations (i) and (ii), we get
$a = 2, b = - 8$

Suggest Corrections

Similar questions

y = a x^{2} + b x

y^{2} = 4 a x

x^{2} = 4 b y

y = x

y = a x^{2} + b x + c

(1, 1)

(- 1, 0)

x^{2} y = c^{3}

a, b

x

y

a^{2} b = (27 / 4) C^{3}

p (x) = x^{3} - a x^{2} + b x + c,

Join BYJU'S Learning Program