Q95 / 160 · AP EAMCET 2022 (04 Jul Shift 1) · Mathematics

The locus of a variable point whose chord of contact w.r.t. the hyperbola x2a2-y2b2=1 subtends a right angle at the origin is

  • A
    x24a2-y24b2=1
  • B
    x2a2-y2b2=x2a4+y2b4
  • C
    xa-yb=1a2+1b2
  • Correct answer
    D
    x2a4+y2b4=1a2-1b2

Solution

Given,

The equation of the given hyperbola is x2a2-y2b2=1     ⋯1

Now let h,k be the pole of a chord PQ of the hyperbola

Then the straight line PQ is the polar of the point h,k w.r.t. the hyperbola and so its equation is

xha2-ykb2=1      ⋯2

Now the center of the hyperbola is the origin, let it be point C,

Now making 1 homogeneous with the help of 2, the combined equation of CP and CQ is

x2a2-y2b2=xha2-ykb22     ⋯3

Now the chord PQ subtends a right angle at 0,0

Therefore the lines CP and CQ given by 3 are perpendicular and so in the equation 3, we have the coefficient of x2+coefficient of y2=0

⇒1a2-h2a4+-1b2-k2b4=0

⇒h2a4+k2b4=1a2-1b2

Thus locus is

x2a4+y2b4=1a2-1b2

Hence, the correct answer is option: 'D'.

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