Home
Class 12
MATHS
Statement 1: The foot of perpendicular f...

Statement 1: The foot of perpendicular from focus on any tangent to ellipse `4x^2 + 5y^2 - 16x + 30y + 41 =0` lie on circle `x^2 + y^2 - 4x - 6y + 4 = 0`. Statement 2: The director circle of ellipse `x^2/a^2 + y^2/b^2 = 1` is `x^2 + y^2 = a^2 + b^2`.

Text Solution

AI Generated Solution

Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

The locus of the foot of the perpendicular from the foci an any tangent to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , is

The centre of the ellipse 4x^(2) + 9y^(2) + 16x - 18y - 11 = 0 is

the foot of perpendicular from a focus on any tangent to the ellipse x^2/4^2 + y^2/3^2 = 1 lies on the circle x^2 + y^2 = 25 . Statement 2: The locus of foot of perpendicular from focus to any tangent to an ellipse is its auxiliary circle.

If the tangent to the ellipse x^2 +4y^2=16 at the point 0 sanormal to the circle x^2 +y^2-8x-4y=0 then theta is equal to

The equation (s) of common tangents (s) to the two circles x^(2) + y^(2) + 4x - 2y + 4 = 0 and x^(2) + y^(2) + 8x - 6y + 24 = 0 is/are

The number of common tangents to the ellipse (x^(2))/(16) + (y^(2))/(9) =1 and the circle x^(2) + y^(2) = 4 is

Find the transverse common tangents of the circles x^(2) + y^(2) -4x -10y + 28 = 0 and x^(2) + y^(2) + 4x - 6y + 4= 0.

Prove that the circles x^(2) +y^(2) - 4x + 6y + 8 = 0 and x^(2) + y^(2) - 10x - 6y + 14 = 0 touch at the point (3,-1)

Statement 1 Feet of prependiculars drawn from foci of an ellipse 4x^(2)+y^(2)=16 on the line 2sqrt3x+y=8 lie on the circle x^(2)+y^(2)=16 Statement 2 If prependiculars are from foci of an ellipse to its any tangent, the feet of these perpendicular lie on director circle of the ellipse.

Find the angle between the circles given by the equations. x^2 + y^2 - 12x - 6y + 41 = 0, x^2 + y^2 + 4x + 6y - 59 = 0.