Medium MCQ +4 / -1 PYQ · JEE Mains 2021

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 :

  1. A (x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $-$ 9x<sup>2</sup> + 16y<sup>2</sup> = 0 Correct answer
  2. B (x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $-$ 9x<sup>2</sup> + 144y<sup>2</sup> = 0
  3. C (x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $-$ 16x<sup>2</sup> + 9y<sup>2</sup> = 0
  4. D (x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $-$ 9x<sup>2</sup> $-$ 16y<sup>2</sup> = 0

Solution

tangent of hyperbola<br><br>$y = mx \pm \sqrt {9{m^2} - 16}$ ..... (i)<br><br>which is a chord of circle with mid-point (h, k)<br><br>so equation of chord T = S<sub>1</sub><br><br>hx + ky = h<sup>2</sup> + k<sup>2</sup><br><br>$y = - {{hx} \over k} + {{{h^2} + {k^2}} \over k}$ ..... (ii)<br><br>by (i) and (ii)<br><br>$m = - {h \over k}$ and $\sqrt {9{m^2} - 16}$ = ${{{h^2} + {k^2}} \over k}$<br><br>$$9{{{h^2}} \over {{k^2}}} - 16 = {{{{\left( {{h^2} + {k^2}} \right)}^2}} \over {{k^2}}}$$<br><br>$\therefore$ Locus : 9x<sup>2</sup> $-$ 16y<sup>2</sup> = (x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup>

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

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