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Q. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4x - 5y = 20$ to the circle $x^2 + y^2 = 9$ is

IIT JEEIIT JEE 2012Conic Sections

Solution:

Plan If $S:ax^2+2hxy+by^2+2gx+2fy+C$
then equation of chord bisected at $P(x_1, y_1)$ is $T=S_1$
or $axx_1+h(xy_1 + yx_1)+byy_1 + g(x+x_1)+f(y+ y_1)+C$
$= ax_1^2+2hx_1y_1 + by^2_1+2gx_1 +2fy, + C$
Description of Situation As equation of chord of contact is $T=0$
image
Here, equation of chord of contact w.r.t. $P$ is
$x\lambda+y\Bigg(\frac{4\lambda-20}{5}\Bigg)=9$
$5\lambda x+ (4\lambda -20)y = 45 ...(i)$
image
and equation of chord bisected a t the point $Q (h , k )$ is
$xh + yk - 9 = h^2 + k^2 - 9$
$\Rightarrow xh + ky = h^2 + k^2 ...(ii)$
From Eqs. (i) and (ii), we get
$ =\frac{5\lambda}{h}=\frac{4\lambda-20}{k}$
$ =\frac{45}{h^2+k^2}$
$\therefore \lambda=\frac{20h}{4h-5k}$
and $ \lambda=\frac{9h}{h^2+k^2}$
$\Rightarrow \frac{20h}{4h - 5k} = \frac{9h}{h^2 + k^2}$
or $ 20 (h^2 + k^2) = 9 (4h - 5k)$
or $20 (x^2+ y^2) = 36 x -45 y$