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

Arun
25750 Points
5 years ago
Dear Rahul

Use mid point of chord of contact formula

T= S1

Hope it helps

Regards
Samyak Jain
333 Points
5 years ago
Let any point on the given circle be O(d cos$\dpi{80} \Theta$, d sin$\dpi{80} \Theta$)
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.
$\dpi{80} \therefore$  dcos$\dpi{80} \Theta$x/a2 + dsin$\dpi{80} \Theta$y/b2 – 1 = 0  i.e.
dcos$\dpi{80} \Theta$x/a2 + dsin$\dpi{80} \Theta$y/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.
$\dpi{80} \therefore$ 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$\dpi{80} \Theta$  =  k/dsin$\dpi{80} \Theta$  =  (h2/a2 + k2/b2) / 1
$\dpi{100} \Rightarrow$ cos$\dpi{80} \Theta$ = h/d(h2/a2 + k2/b2)  &  sin$\dpi{80} \Theta$ = k/d(h2/a2 + k2/b2)

$\dpi{100} \because$ cos2$\dpi{80} \Theta$ + sin2$\dpi{80} \Theta$ = 1
$\dpi{100} \therefore$ [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.
$\dpi{100} \therefore$   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.