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Hyperbola question

2021 · 16 Mar · Shift 1 · Q30
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  5. /2021 · 16 Mar · Shift 1 · Q30

Hyperbola question

2021 · 16 Mar · Shift 1 · Q30

JEE MainMathematicsHyperbolaMCQ+4 / −1
The locus of the midpoints of the chord of the circle, x2 + y2 = 25 which is tangent to the hyperbola, x29−y216=1{{{x^2}} \over 9} - {{{y^2}} \over {16}} = 19x2​−16y2​=1 is :
  1. A
    (x2 + y2)2 −-− 9x2 + 16y2 = 0
  2. B
    (x2 + y2)2 −-− 9x2 + 144y2 = 0
  3. C
    (x2 + y2)2 −-− 16x2 + 9y2 = 0
  4. D
    (x2 + y2)2 −-− 9x2 −-− 16y2 = 0
View written solutionFree

Correct answer: A

  1. Let the midpoint of the chord be (h,k)(h,k)(h,k).

    For the circle x2+y2=25,x^2+y^2=25,x2+y2=25, the chord whose midpoint is (h,k)(h,k)(h,k) has equation xh+yk=h2+k2.xh+yk=h^2+k^2.xh+yk=h2+k2.

    This is the standard chord-with-given-midpoint form for the circle x2+y2=a2x^2+y^2=a^2x2+y2=a2.

  2. This chord is tangent to the hyperbola x29−y216=1.\frac{x^2}{9}-\frac{y^2}{16}=1.9x2​−16y2​=1.

    So the line xh+yk=h2+k2xh+yk=h^2+k^2xh+yk=h2+k2 must be a tangent to the hyperbola.

  3. Use the tangent condition for the hyperbola.

    For the hyperbola x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,a2x2​−b2y2​=1, a line lx+my=nlx+my=nlx+my=n is tangent iff n2=a2l2−b2m2.n^2=a^2l^2-b^2m^2.n2=a2l2−b2m2.

    Here, l=h,m=k,n=h2+k2,a2=9,b2=16.l=h,\quad m=k,\quad n=h^2+k^2,\quad a^2=9,\quad b^2=16.l=h,m=k,n=h2+k2,a2=9,b2=16.

    Therefore, (h2+k2)2=9h2−16k2.(h^2+k^2)^2=9h^2-16k^2.(h2+k2)2=9h2−16k2.

  4. Hence the locus of (h,k)(h,k)(h,k) is (h2+k2)2−9h2+16k2=0.(h^2+k^2)^2-9h^2+16k^2=0.(h2+k2)2−9h2+16k2=0.

    Replacing (h,k)(h,k)(h,k) by (x,y)(x,y)(x,y), the locus is (x2+y2)2−9x2+16y2=0.(x^2+y^2)^2-9x^2+16y^2=0.(x2+y2)2−9x2+16y2=0.

  5. Match with the options.

    This is exactly Option A: (x2+y2)2−9x2+16y2=0.(x^2+y^2)^2-9x^2+16y^2=0.(x2+y2)2−9x2+16y2=0.


Verification with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

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