Find the locus of the midpoint of the chord o | vedclass

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(C) The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$. The equation of a tangent to this hyperbola with slope $m$ is $y = mx \pm \

Vedclass · Accepted Answer

${\left( {{x^2} + {y^2}} \right)^2} = 16x^2 - 9y^2$

Vedclass · Answer

(C) The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$. The equation of a tangent to this hyperbola with slope $m$ is $y = mx \pm \sqrt{16m^2 - 9} \dots (i)$.Let $(x_1, y_1)$ be the midpoint of the chord of the circle $x^2 + y^2 = 16$. The equation of the chord with midpoint $(x_1, y_1)$ is given by $T = S_1$,which is $xx_1 + yy_1 = x_1^2 + y_1^2 \dots (ii)$.Comparing $(i)$ and $(ii)$,we rewrite $(ii)$ as $y = -\frac{x_1}{y_1}x + \frac{x_1^2 + y_1^2}{y_1}$.Equating the slopes and intercepts: $m = -\frac{x_1}{y_1}$ and $\sqrt{16m^2 - 9} = \frac{x_1^2 + y_1^2}{y_1}$.Squaring both sides: $16m^2 - 9 = \frac{(x_1^2 + y_1^2)^2}{y_1^2}$.Substituting $m = -\frac{x_1}{y_1}$: $16\left(-\frac{x_1}{y_1}\right)^2 - 9 = \frac{(x_1^2 + y_1^2)^2}{y_1^2}$.$16x_1^2 - 9y_1^2 = (x_1^2 + y_1^2)^2$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 = 16x^2 - 9y^2$.

Find the locus of the midpoint of the chord of the circle $x^2 + y^2 = 16$ which is a tangent to the hyperbola $9x^2 - 16y^2 = 144$.

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