Question

The point of intersection of the perpendicular tangents to the ellipse  lies on a circle square of whose radius is

 

Solution

Correct option is

2906

Any tangent to the ellipse is

         and perpendicular to it is

        

Eliminating m we get the locus of the point of intersection of these tangents.

        (y – mx)2 + (mx + x)2 = (1 + m2)[(41)2 + (35)2]

     x2 + y2 = (41)2 + (35)2 which is a circle square of whose radius = (41)2 + (35)2 = 1681 + 1225 = 2906.

SIMILAR QUESTIONS

Q1

P is a point on the circle C, the perpendicular PQ to the major axis of the ellipse E meets the ellipse at M, then  is equal to

Q2

Equation of the diameter of the ellipse conjugate to the diameter respected by L is 

Q3

 

If R is the point of intersection of the line L with the line x = 1, then 

Q4

If L is the chord of contact of the hyperbola H, then the equation of the corresponding pair of tangents is

Q5

If R is the point of intersection of the tangents to H at the extremities  of the chord L, then equation of the chord contact of with respect to the parabola P is

Q6

If the chord of contact of R with respect to the parabola Pmeets the parabola at T and T’, S is the focus of the parabola, then Area of the triangle STT’ is equal to

Q7

Tangent are drawn from any point on the hyperbola  to the circle x2 + y2 = 9. If the locus of the mid-point of the chord of contact is

  

Q8

If l is the length of the intercept made by a common tangent to the circle x2 + y= 16 and the ellipse , on the coordinate axes, then 81l2+ 3 is equal to

Q9

If d is the length of the tangent from the point (100, 81) to the ellipse , then d2 is equal to

Q10

If x + y = k is a normal to the parabola y2 = 12xp is the length of the perpendicular from the focus of the parabola on this normal; then 3k2 + 2p2 is equal to