Home
Class 12
MATHS
Let P be the point on parabola y^2=4x wh...

Let P be the point on parabola `y^2=4x` which is at the shortest distance from the center S of the circle `x^2+y^2-4x-16y+64=0` let Q be the point on the circle dividing the line segment SP internally. Then

A

`SP = 2 sqrt(5)`

B

`SQ : QP (sqrt(5) + 1) : 2`

C

the x-intercept of the normal to the parabola at P is 6

D

the slope of the tangent to the circle at Q is `(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D

Tangent to `y^(2) = 4x` at `(t^(2), 2t)` is

`y(2 t) = 2(x + t^(2))`
`implies yt = x + t^(2)`
Equation of normal of `P(t^(2) , 2t)` is
`y + tx = 2t + t^(3)`
Since, normal at P passes through centre of circle S(2,8).
`:. 8 + 2 t + t^(3)`
`implies t = 2,` i., P (4,4)
[Since, shortest distance between two curves lie along their common normal and the common normal will pass through the centre of circle]
`:. SP = sqrt((4 - 2)^(2) + (4 - 8)^(2)) = 2 sqrt(5)`
`:.` Opion (a) is correct.
Also, `SQ = 2`
`:. PQ = SP - SQ = 2 sqrt(5) - 2`
Thus, `(SQ)/(QP) = (1)/(sqrt(5) - 1) = (sqrt(5) + 1)/(4)`
`:.` Option (b) is wrong
Now x intercept of normal is `x = 2 + 2^(2) = 6`
`:.` Option (c ) is corect
Slope of tangent `= (1)/(t) = (1)/(2)`
`:.` Option (d) is correct.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • PARABOLA

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 3 EQUATION OF NORMAL AND PROPERTIES OBJECTIVES QUESTIONS II (INTEGER )|1 Videos
  • PARABOLA

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 3 EQUATION OF NORMAL AND PROPERTIES OBJECTIVES QUESTIONS II (ANALYTICAL & DESCRIPTIVE )|4 Videos
  • PARABOLA

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 3 EQUATION OF NORMAL AND PROPERTIES OBJECTIVES QUESTIONS I (MATCH THE COLUMNS )|1 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

Let P be the point on the parabola y^(2) = 4x which is at the shortest distance from the centre S of the circle x^(2) + y^(2) - 4x - 16y + 64 = 0 . Let Q be the point on the circle dividing the lie segment SP internally. Then

Find the point on y=x^2+4 which is at shortest distance from the line y=4x-4

Knowledge Check

  • Let P be a point on the parabola, x^2=4y . If the distance of P from the centre of the circle, x^2+y^2+6x+8=0 is minimum, then the equation of the tangent to the parabola at P, is:

    A
    `x+4y-2=0`
    B
    `x+y+1=0`
    C
    `x-y+3=0`
    D
    `x+2y=0`
  • Similar Questions

    Explore conceptually related problems

    The shortest distance from (-2,14) to the circle x^(2)+y^(2)-6x-4y-12=0 is

    Let P be the point on the parabola, y^2 = 8x which is at a minimum distance from the center C of the circle x^2 + (y+6)^2 = 1 . Then the equation of the circle, passing through C and having its center at P is

    The shortest distance from the line 3x+4y=25 to the circle x^(2)+y^(2)=6x-8y is equal to :

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

    Let P be the point on the parabola, y^2=8x which is at a minimum distance from the centre C of the circle,x^2+(y+6)^2=1. Then the equation of the circle, passing through C and having its centre at P is : (1) x^2+y^2-4x+8y+12=0 (2) x^2+y^2-x+4y-12=0 (3) x^2+y^2-x/4+2y-24=0 (4) x^2+y^2-4x+9y+18=0

    Find the greatest distance of the point P(10,7) from the circle x^(2)+y^(2)-4x-2y-20=0

    If the shortest distance of the parabola y^(2)=4x from the centre of the circle x^(2)+y^(2)-4x-16y+64=0 is "d" ,then d^(2) is equal to: