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