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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

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

Grade:12th pass

2 Answers

Arun
25750 Points
5 years ago
Dear Rahul
 
Use mid point of chord of contact formula
 
T= S1
 
Hope it helps
 
Regards
Arun (askIITians forum expert)
Samyak Jain
333 Points
5 years ago
Let any point on the given circle be O(d cos\Theta, d sin\Theta)
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  dcos\Thetax/a2 + dsin\Thetay/b2 – 1 = 0  i.e. 
dcos\Thetax/a2 + dsin\Thetay/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/dcos\Theta  =  k/dsin\Theta  =  (h2/a2 + k2/b2) / 1
\Rightarrow cos\Theta = h/d(h2/a2 + k2/b2)  &  sin\Theta = k/d(h2/a2 + k2/b2)
 
\because cos2\Theta + sin2\Theta = 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)2 is the required locus.
(You may simplify the above equation to get the desired result.)
Pls approve.

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