QJEEM21Ma5c12: JEE Main 2021 Mathematics - Class 11 JEEM21Ma5c12 Solution: The locus of the midpoints of the chord of the circle, x 2 + y…

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The locus of the midpoints of the chord of the circle, x² + y² = 25 which is tangent to the hyperbola,

{{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1

is :

Accepted Answer

tangent of hyperbola

y = mx \pm \sqrt {9{m^2} - 16}

..... (i)

which is a chord of circle with mid-point (h, k)

so equation of chord T = S₁

hx + ky = h² + k²

y = - {{hx} \over k} + {{{h^2} + {k^2}} \over k}

..... (ii)

by (i) and (ii)

m = - {h \over k}

and

\sqrt {9{m^2} - 16}

=

{{{h^2} + {k^2}} \over k}

9{{{h^2}} \over {{k^2}}} - 16 = {{{{\left( {{h^2} + {k^2}} \right)}^2}} \over {{k^2}}}

\therefore

Locus : 9x²

-

16y² = (x² + y²)²

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