The locus of the midpoints of the chords of t | vedclass

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(D) Let the midpoint of the chord be $(h, k)$. The equation of the chord of the circle $x^2 + y^2 = a^2$ with midpoint $(h, k)$ is given by $T = S_1$,

Vedclass · Accepted Answer

$(x^2 + y^2)^2 = a^2x^2 - b^2y^2$

Vedclass · Answer

(D) Let the midpoint of the chord be $(h, k)$. The equation of the chord of the circle $x^2 + y^2 = a^2$ with midpoint $(h, k)$ is given by $T = S_1$,which is $xh + yk = h^2 + k^2$.This can be rewritten as $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$.Comparing this with the line $y = mx + c$,we have $m = -\frac{h}{k}$ and $c = \frac{h^2 + k^2}{k}$.Since this chord touches the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition for tangency $c^2 = a^2m^2 - b^2$ must be satisfied.Substituting the values of $m$ and $c$:$\left(\frac{h^2 + k^2}{k}\right)^2 = a^2\left(-\frac{h}{k}\right)^2 - b^2$$(h^2 + k^2)^2 = a^2h^2 - b^2k^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 = a^2x^2 - b^2y^2$.

The locus of the midpoints of the chords of the circle $x^2 + y^2 = a^2$ which touch the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

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