Home
Class 12
MATHS
The vertex of the parabola (y-1)^2 =8 (x...

The vertex of the parabola `(y-1)^2 =8 (x-1)` is at the centre of a circle and the parabola cuts that circle at the ends of its latusrectum. Then the equation of that circle is

Text Solution

Verified by Experts

The correct Answer is:
`x^(2)+y^(2)=20`
Promotional Banner

Topper's Solved these Questions

  • PARABOLA

    AAKASH SERIES|Exercise EXERCISE- I|57 Videos
  • PARABOLA

    AAKASH SERIES|Exercise EXERCISE- II|100 Videos
  • PARABOLA

    AAKASH SERIES|Exercise EXERCISE - 3.3 ( SHORT ANSWER QUESTIONS )|8 Videos
  • MEASURES OF DISPERSION (STATISTICS)

    AAKASH SERIES|Exercise Practice Exercise|54 Videos
  • PARTIAL FRACTIONS

    AAKASH SERIES|Exercise PRACTICE EXERCISE|31 Videos

Similar Questions

Explore conceptually related problems

The vertex of the parabola x^(2)+8x+12y+4=0 is

Find the equation of tangent to the parabola y^(2)=4x at the end of latusrectum in the first quadrant

The equation of the normal at the end of latusrectum in the fourth quadrant of the parabola y^(2)=4ax is

The ends of latusrectum of parabola are (4, 8) and (4,-8) then equation of parabola is

If a circle cuts a parabola in four points then the sum of ordinates of four points is

The vertex of a parabola is the point (a,b) and latusrectum is of length 1. If the axis of the parabola is along the positive direction of y-axis then its equation is