Home
Class 12
MATHS
Circle of constant radius r are draw to ...

Circle of constant radius r are draw to pass through the ends of a variable diameter of the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1`. Prove that locus of their centres is the curve
`(x^(2)+y^(2))(a^(2)x^(2)+b^(2)y^(2)+a^(2)b^(2))=r^(2)(a^(2)x^(2)+b^(2)y^(2))`

Text Solution

Verified by Experts

The correct Answer is:
(h,k).
Promotional Banner

Topper's Solved these Questions

  • CO-ORDINATE GEOMETRY OF THREE DIMENSION

    ML KHANNA|Exercise SELF ASSIGNMENT TEST |11 Videos
  • CO-ORDINATE GEOMETRY OF THREE DIMENSION

    ML KHANNA|Exercise Assertion/Reason|4 Videos
  • BINOMIAL THEOREM AND MATHEMATICAL INDUCTION

    ML KHANNA|Exercise Self Assessment Test |35 Videos
  • COMPLEX NUMBERS

    ML KHANNA|Exercise Assertion / Reason |2 Videos

Similar Questions

Explore conceptually related problems

The area of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

Chords of the circle x^(2)+y^(2)=a^(2) toches the hyperbola x^(2)/a^(2)-y^(2)//b^(2)=1 . Prove that locus of their middle point is the curve (x^2 +y^2)^2=a^(2)x^(2)-b^(2)y^(2)

Tangents at right angle are drawn to the ellipse x^(2)/a^(2)+y^(2)/b^(2)=1 . Show that the focus of the middle points of the chord of contact is the curve (x^(2)/a^(2)+y^(2)/b^(2))^(2)=(x^(2)+y^(2))/(a^(2)+b^(2)) .

Centre of the ellipse ((x-h)^(2))/(a^(2))+((y-k)^(2))/(b^(2))=1(a

Show that the tangents at the ends of conjugate diameters of the ellipse x^(2)/a^(2)+y^(2)/b^(2)=1 intersect on the ellipse x^(2)/a^(2)+y^(2)/b^(2)=2 .

The positive end of the Latusrectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1(a>b)" is

the equation of a diameter conjugate to a diameter y=(b)/(a)x of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

For the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 and (x^(2))/(b^(2))+(y^(2))/(a^(2)) =1

The locus of the point of intersection of tangents at the end-points of conjugate diameters of the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , is