Home
Class 12
MATHS
The points(s) on the curve y^3+3x^2=12 y...

The points(s) on the curve `y^3+3x^2=12 y` where the tangent is vertical is (are) `(+-4/(sqrt(3)),-2)` (b) `(+-sqrt((11)/3),1)` `(0,0)` (d) `(+-4/(sqrt(3)),2)`

A

`(pm (4)/(sqrt3), -2)`

B

`(pm sqrt((11)/(3)), 0)`

C

`(0, 0`)

D

` (pm (4)/(sqrt3), 2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given, `" " y^(3) + 3x ^(2) = 12 y " " `…(i)
On differentiating w.r.t. x, we get
`rArr " " 3y^(2) (dy)/(dx) + 6x = 12 (dy)/(dx)`
`rArr " " (dy)/(dx) = ( 6x)/(12 - 3y^(2))`
`rArr " " (dx)/(dy) = ( 12 - 3y^(2))/(6x)`
For vertical tangent, `(dx)/(dy) =0`
`rArr 12 - 3y^(2) =0 rArr y = pm 2`
On putting, `y = 2` in Eq. (i), we get ` x = pm ( 4)/(sqrt3)` and again putting `y = - 2 ` in Eq. (i), we get ` 3x^(2) = - 16`, no real solution .
So, the required point is `(pm (4)/(sqrt(3)), 2)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

The point(s) on the curve y^3+\ 3x^2=12 y where the tangent is vertical, is(are) ?? (+-4/(sqrt(3)),\ -2) (b) (+-\ sqrt((11)/3,\ )\ 1) (0,\ 0) (d) (+-4/(sqrt(3)),\ 2)

Points on the curve f(x)=x/(1-x^2) where the tangent is inclined at an angle of pi/4 to the x-axis are (a)(0,0) (b) (sqrt(3),-(sqrt(3))/2) (-2,2/3) (d) (-sqrt(3),(sqrt(3))/2)

A is a point on either of two lines y+sqrt(3)|x|=2 at a distance of 4/sqrt(3) units from their point of intersection. The coordinates of the foot of perpendicular from A on the bisector of the angle between them are (a) (-2/(sqrt(3)),2) (b) (0,0) (c) (2/(sqrt(3)),2) (d) (0,4)

The incenter of the triangle with vertices (1,sqrt(3)),(0,0), and (2,0) is (a) (1,(sqrt(3))/2) (b) (2/3,1/(sqrt(3))) (c) (2/3,(sqrt(3))/2) (d) (1,1/(sqrt(3)))

A beam of light is sent along the line x-y=1 , which after refracting from the x-axis enters the opposite side by turning through 30^0 towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is (a) (2-sqrt(3))x-y=2+sqrt(3) (b) (2+sqrt(3))x-y=2+sqrt(3) (c) (2-sqrt(3))x+y=(2+sqrt(3)) (d) y=(2-sqrt(3))(x-1)

If the area enclosed between the curves y=kx^2 and x=ky^2 , where kgt0 , is 1 square unit. Then k is: (a) 1/sqrt(3) (b) sqrt(3)/2 (c) 2/sqrt(3) (d) sqrt(3)

The vertices of a triangle are (0,0), (x ,cosx), and (sin^3x ,0),w h e r e0ltxltpi/2 the maximum area for such a triangle in sq. units is (a) (3sqrt(3))/(32) (b) (sqrt(3)/32) (c) 4/32 (d) (6sqrt(3)) /(32)

The value of cot70^0+4cos70^0 is (a) 1/(sqrt(3)) (b) sqrt(3) (c) 2sqrt(3) (d) 1/2

Tangent of acute angle between the curves y=|x^2-1| and y=sqrt(7-x^2) at their points of intersection is (5sqrt(3))/2 (b) (3sqrt(5))/2 (5sqrt(3))/4 (d) (3sqrt(5))/4

Find the point on the curve where tangents to the curve y^2-2x^3-4y+8=0 pass through (1,2).