JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 8)
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 :
(x2 + y2)2 $$-$$ 9x2 + 16y2 = 0
(x2 + y2)2 $$-$$ 9x2 + 144y2 = 0
(x2 + y2)2 $$-$$ 16x2 + 9y2 = 0
(x2 + y2)2 $$-$$ 9x2 $$-$$ 16y2 = 0
Explanation
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
$$y = mx \pm \sqrt {9{m^2} - 16} $$ ..... (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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