Locus of the mid points of the chord of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, so that chord is always touching the circle $x^{2} + y^{2} = c^{2}$, $(c

Question

Locus of the mid points of the chord of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, so that chord is always touching the circle $x^{2} + y^{2} = c^{2}$, $(c < a,\ c < b)$ is

Dr. Abubacker · Accepted Answer

(a)
Let mid point of the chord be $(h,\ k)$, then equation of the chords be
$\frac{hx^{2}}{a^{2}} + \frac{{ky}^{2}}{b^{2}} - 1 = \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} - 1$
$\Rightarrow y = - \frac{b^{2}}{a^{2}} \bullet \frac{h}{k}x + \left( \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} ight)\frac{b^{2}}{k}$ …(i)
Since, line (i) is touching the circle $x^{2} + y^{2} = c^{2}$
$	herefore\left( \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} ight)\frac{b^{4}}{k^{2}} = c^{2}\left( 1 + \frac{b^{4}h^{2}}{a^{4}k^{2}} ight)$
Hence, locus is $\left( b^{2}x^{2} + a^{2}y^{2} ight)^{2} = c^{2}(b^{4}x^{2} + a^{4}y^{2})$

Q. Locus of the mid points of the chord of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, so that chord is always touching the circle $x^{2} + y^{2} = c^{2}$, $(c < a,\ c < b)$ is

Explanation

Reviewed by Dr. Abubacker

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