The distance between the pair of lines \(x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0\), is

The distance between the pair of lines \(x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0\), is
  1. \(4 \sqrt{2}\)
  2. \(2 \sqrt{2}\)
  3. 2
  4. \(6 \sqrt{2}\)

Solution

Given, equation of pair of lines is \(x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0\) The distance between the pair of lines \(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\) is given by \(d=2 \sqrt{\frac{g^2-a c}{a(a+b)}}\) Here, \(a=1, b=2, c=1, g=2\) \(\therefore\) Required distance \(=2 \sqrt{\frac{4-1}{1(1+2)}}=2 \sqrt{\frac{3}{3}}=2\)

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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