Question

From afrom a point o on the circle x^2 +y^2=d^2,tangent op and oq are drawn to the ellipse x^2/a^2+y^2/b^2=1.Find locus of the midpoint of the chord pq

Answer 1

Here u can easily get answer by using the formula T = S1

Answer of this question is

x^2+y^2 = d^2[(x/a)^2 + (y/b)^2]^2

Answer 2

Answer: x² + y² = d²(x²/a² + y²/b²)

Explanation:Let any point on the given circle be O(d cosθ, d sinθ)

and midpoint of chord PQ of ellipse be M(h,k).

Equation of chord of contact PQ with O as an external point is

T = 0 w.r.t. the ellipse x²/a² + y²/b² – 1 = 0.

therefore dcosθx/a² + dsinθy/b² – 1 = 0 i.e.

dcosθx/a² + dsinθy/b² = 1 ….(1) is the equation of PQ.

Also, considering midpoint M, equation of chord PQ is

T = S1 w.r.t. the ellipse x²/a² + y²/b² – 1 = 0.

\therefore hx/a² + ky/b² – 1 = h²/a² + k²/b² – 1 i.e.

hx/a² + ky/b² = h²/a² + k²/b² ….(2) is the equation of PQ.

Comparing (1) & (2), we get

h/dcosθ = k/dsinθ = (h²/a² + k²/b²) / 1

cosθ = h/d(h²/a² + k²/b²) & sinθ = k/d(h²/a² + k²/b²)

because cos²θ + sin²θ = 1

therefore [h/d(h²/a² + k²/b²)]² + [k/d(h²/a² + k²/b²)]² = 1

h² + k² = d²(h²/a² + k²/b²)²

Replace h by x and k by y.

Therefore x² + y² = d²(x²/a² + y²/b²) is the required locus.