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AP EAMCET · Maths · Straight Lines

The transformed equation of \(3 x^2-6 x y+8 y^2=8\) when the axes are rotated about the origin through an angle \(\frac{\pi}{4}\) in the positive direction, is

  1. A \(5 x^2+10 x y+17 y^2+16=0\)
  2. B \(5 x^2+10 x y+17 y^2-16=0\)
  3. C \(5 x^2-10 x y+17 y^2-16=0\)
  4. D \(5 x^2-10 x y+17 y^2+16=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(5 x^2+10 x y+17 y^2-16=0\)

Step-by-step Solution

Detailed explanation

Given transformed equation is \[ 3 x^2-6 x y+8 y^2=8 \] Now, \(x^{\prime}=x \cos \frac{\pi}{4}-y \sin \frac{\pi}{4}=\frac{x-y}{\sqrt{2}}\) and \(y^{\prime}=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}}\) Before transformation the equation is…