Home
Class 12
MATHS
Given : A circle, 2x^2+""2y^2=""5 and a ...

Given : A circle, `2x^2+""2y^2=""5` and a parabola, `y^2=""4sqrt(5)""x` .

A

Statement I is correct, Statement II is correct, Statement II is a correct explanation for statement I

B

Statement I is correct, Statement II is correct, Statement II is not a correct explanation for statement I

C

Statement I is correct, Statement II is incorrect

D

Statement I is incorrect, Statement II is correct

Text Solution

Verified by Experts

The correct Answer is:
A
Equation of circle can be rewriteen as `x^(2) + y^(2) = (5)/(2)`
Centre `to` (0,0) and radius `to sqrt((5)/(2))`
Let common tangent be
`y = mx + (sqrt(5))/(m) implies m^(2) x - my + sqrt(5) = 0`
The perpendicular from centre to the tangenet is equal to radius of the circle.
`:. (sqrt(5)//m)/(sqrt(1 + m^(2))) = sqrt((5)/(2))`
`implies m sqrt(1 + m^(2)) = sqrt(2)`
`implies m^(2) (1 + m^(2)) = 2`
`implies (m^(2) + 2) (m^(2) = 1) = 0`
`implies m = +- 1` `[:' m^(2) + 2 =! 2 =!` as `m in R]`
`:. y = +- (x + sqrt(5))`, both statements are correct as `m +- 1`
satisfies the given equation of statement II.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • PARABOLA

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 2 EQUATION OF TANGENTS AND PROPERTIES OBJECTIVES QUESTIONS II (ONE OR MORE THAN CORRECT OPTION )|1 Videos

  • PARABOLA

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 2 EQUATION OF TANGENTS AND PROPERTIES (PASSAGE BASED PROBLEMS )|2 Videos

  • PARABOLA

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 2 EQUATION OF TANGENTS AND PROPERTIES OBJECTIVE QUESTIONS I (ONLY ONE CORRECT OPTION )|13 Videos

  • MISCELLANEOUS

    IIT JEE PREVIOUS YEAR|Exercise MISCELLANEOUS|87 Videos

  • PERMUTATIONS AND COMBINATIONS

    IIT JEE PREVIOUS YEAR|Exercise Dearrangement and Number of Divisors (Fill in the Blank )|1 Videos

Similar Questions

Explore conceptually related problems

Given : A circle 2x^2 + 2y^2 = 5 and a parabola y^2 = 4sqrt5 x Statement-1 : An equation of the common tangent to these curve is y = x + sqrt5 Statement-2 : If the line y = mx + (sqrt5)/m,mne0 is their common tangent, then m satisfies m^4 - 3m^2 + 2 = 0

Two common tangents to the circle x^(2)+y^(2)=a^(2) and parabola y^(2)=4a sqrt(2)x are

Knowledge Check

  • The equation of the line which is tangent to both the circle x^2+y^2=5 and the parabola y^2=40x is

    A
    `+-2x-y+-5=0`
    B
    2x-y+5=0
    C
    2x-y-5=0
    D
    2x+y+5=0

  • Equation of a common tangent to the circle x^(2)+y^(2)-6x=0 and the parabola y^(2)=4x is

    A
    `2sqrt(3) y=12x+1`
    B
    `2sqrt(3)y=-x-12`
    C
    `sqrt(3)y=x+3`
    D
    `sqrt(3)y=3x+1`

  • Equation of a common tangent to the circle x^(2) + y^(2) - 6x = 0 and the parabola, y^(2) = 4x , is

    A
    `sqrt(3) y = 3x + 1`
    B
    `2 sqrt(3) y = 12x + 1`
    C
    `sqrt(3) y = x + 3`
    D
    `2 sqrt(3) y = - x - 12`

    • Similar Questions

    Explore conceptually related problems

    Area common to the circle x^(2)+y^(2)=64 and the parabola y^(2)=4x is

    The equation(s) of the line(s) which touch both the circle x^(2)+y^(2)=5 and the parabola y^(2)=40x is

    Shortest distance between the centre of circle x^2+y^2-4x-16y+64=0 and parabola y^2=4x is:

    Statement 1: There are no common tangents between the circle x^(2)+y^(2)-4x+3=0 and the parabola y^(2)=2x. Statement 2: Given circle and parabola do not intersect.

    Two common tangents to the circle x^(2) + y^(2) = (a^(2))/(2) and the parabola y^(2) = 4ax are

    Home
    Profile