Home
Class 12
MATHS
The equation of the common tangent touch...

The equation of the common tangent touching the circle `(x-3)^2+y^2=9` and the parabola `y^2=4x` above the x-axis is `sqrt(3)y=3x+1` (b) `sqrt(3)y=-(x+3)` `sqrt(2)y=x+3` (d) `sqrt(3)y=-(3x-1)`

A

`sqrt3y=3x+1`

B

`sqrt3y=-(x+3)`

C

`sqrt3y=x+3`

D

`sqrt3y=-(3x+1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`y^(2)=4" "(x-3)^(2)+y^(2)=9`
`y=mx+(1)/(m)" "(3,0),r=3`
`|(3m+(1)/(m))/(sqrt(1+m^(2)))|=3`
`9m^(2)+(1)/(m^(2))+6=9+9m^(2)`
`(1)/(m^(2))=3impliesm= pm(1)/(sqrt3)`
`m=(1)/(sqrt3)` in first quadrant
`y=(x)/(sqrt3)+sqrt3impliessqrt3y=x+3`
Promotional Banner

Topper's Solved these Questions

  • MOCK TEST 1

    VMC MODULES ENGLISH|Exercise PART III : MATHEMATICS (SECTION-2)|10 Videos

  • MATRICES AND DETERMINANTS

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ARCHIVE|78 Videos

  • MOCK TEST 10

    VMC MODULES ENGLISH|Exercise MATHEMATICS (SECTION 2)|5 Videos

Similar Questions

Explore conceptually related problems

The equation of the common tangent touching the circle (x-3)^2+y^2=9 and the parabola y^2=4x above the x-axis is sqrt(3)y=3x+1 (b) sqrt(3)y=-(x+3) (C) sqrt(3)y=x+3 (d) sqrt(3)y=-(3x-1)

The equation of common tangent to the parabola y^2 =8x and hyperbola 3x^2 -y^2=3 is

The slopes of the common tangents of the ellipse (x^2)/4+(y^2)/1=1 and the circle x^2+y^2=3 are +-1 (b) +-sqrt(2) (c) +-sqrt(3) (d) none of these

Find the equation of the tangent line to the curve y=sqrt(5x-3)-2 which is parallel to the line 4x-2y+3=0 .

The equation of the tangents to the ellipse 4x^2 + 3y^2 = 5 , which are inclined at 60^0 to the X-axis are : (A) y = x/sqrt(3) +- sqrt(65/12) (B) y = sqrt(3) x +- sqrt(65/12) (C) y = sqrt(3)x +- sqrt(12/65) (D) none of these

If 5-sqrt(3)=x+y sqrt(3) then (x,y) is

Find the equation of the tangent line to the curve y=sqrt(5x-3)-2 which is parallel to the line 4x-2y+3=0

The equation of the lines passing through the point (1,0) and at a distance (sqrt(3))/2 from the origin is (a) sqrt(3)x+y-sqrt(3) =0 (b) x+sqrt(3)y-sqrt(3)=0 (c) sqrt(3)x-y-sqrt(3)=0 (d) x-sqrt(3)y-sqrt(3)=0

The area of the circle x^2+y^2=16 exterior to the parabola y^2=6x is(A) 4/3(4pi-sqrt(3)) (B) 4/3(4pi+sqrt(3)) (C) 4/3(8pi-sqrt(3)) (D) 4/3(8pi+sqrt(3))

Prove that the lines sqrt(3)x+y=0,\ sqrt(3)y+x=0,\ sqrt(3)x+y=1\ a n d\ sqrt(3)y+x=1 form a rhombus.