If the curve $y^{2} = a x^{3} - 6 x^{2} + b$ passes through $(0, 1)$ and has its tangent parallel to y-axis at $x = 2$ , then

a = 2, b = 1

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

a = \frac{23}{8}, b = 1

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

a = - \frac{8}{23}, b = 1

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

a = - \frac{23}{8}, b = 1

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

Open in App

Solution

The correct option is A $a = \frac{23}{8}, b = 1$
Given equation of curve
$y^{2} = a x^{3} - 6 x^{2} + b$
Since, the curve passes through $(0, 1)$
$\Rightarrow b = 1$
Since, the tangent is parallel to y-axis at $x = 2$
$\Rightarrow y = 8 a - 23$
So, point of tangency is $(2, 8 a - 23)$
Slope of tangent to curve
$2 y \frac{d y}{d x} = 3 a x^{2} - 12 x$
$\Rightarrow \frac{d y}{d x} = \frac{3 a x^{2} - 12 x}{2 y}$
Slope of tangent to curve at $(2, 8 a - 23)$ is $\frac{12 a - 24}{16 a - 46}$
Since, the tangent is parallel to y-axis
$\frac{12 a - 24}{16 a - 46} = \frac{1}{0}$
$\Rightarrow 16 a - 46 = 0$
$\Rightarrow a = \frac{23}{8}$

Suggest Corrections

Similar questions

y = a x^{2} + b x

y = a \sqrt{x} + b x

$3 + 23 y - 8 y^{2} = ? (a) (1 - 8 y) (3 + y) (b) (1 + 8 y) (3 - y) (c) (1 - 8 y) (y - 3) (d) (8 y - 1) (y + 3)$

y = f (x)

P (x, y)

A P : P B = 1 : 3

f (1) = 1

(k, \frac{1}{8})

k

Join BYJU'S Learning Program