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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2021 · 16 March · Morning Shift · Mathematics · Hyperbola

The locus of the midpoints of the chord of the circle, x ^ 2 + y ^...

The locus of the midpoints of the chord of the circle, x2+y2=25{x}^{2} + {y}^{2} = 25 which is tangent to the hyperbola, x29−y216=1\dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{16} = 1 is :The locus of the midpoints of the chord of the circle, x ^ 2 + y ^ 2 = 25 which is tangent to the hyperbola, x ^ 2 9 -

Options

  1. A(x2+y2)2−16x2+9y2=0{\left({x}^{2} + {y}^{2}\right)}^{2} - 16{x}^{2} + 9{y}^{2} = 0
  2. B(x2+y2)2−9x2+144y2=0{\left({x}^{2} + {y}^{2}\right)}^{2} - 9{x}^{2} + 144{y}^{2} = 0
  3. C(x2+y2)2−9x2−16y2=0{\left({x}^{2} + {y}^{2}\right)}^{2} - 9{x}^{2} - 16{y}^{2} = 0
  4. D(x2+y2)2−9x2+16y2=0{\left({x}^{2} + {y}^{2}\right)}^{2} - 9{x}^{2} + 16{y}^{2} = 0

Correct answer

D. (x2+y2)2−9x2+16y2=0{\left({x}^{2} + {y}^{2}\right)}^{2} - 9{x}^{2} + 16{y}^{2} = 0

Step-by-step solution

The locus of the midpoints of the chord of the circle, x ^ 2 + y ^ 2 = 25 which is tangent to the hyperbola, x ^ 2 9 -

Equation of chord

y−k=−hk(x−h)y - k = - \dfrac{h}{k}\left(x - h\right)

ky−k2=−hx+h2ky - {k}^{2} = - hx + {h}^{2}

hx+ky=h2+k2hx + ky = {h}^{2} + {k}^{2}

y=−hxk+h2+k2ky = - \dfrac{hx}{k} + \dfrac{{h}^{2} + {k}^{2}}{k}

tangent to x29−y216=1\dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{16} = 1

c2=a2m2−b2{c}^{2} = {a}^{2}{m}^{2} - {b}^{2}

(h2+k2k)2=9(−hk)2−16{\left(\dfrac{{h}^{2} + {k}^{2}}{k}\right)}^{2} = 9{\left(- \dfrac{h}{k}\right)}^{2} - 16

(x2+y2)2=9x2−16y2{\left({x}^{2} + {y}^{2}\right)}^{2} = 9{x}^{2} - 16{y}^{2}

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