JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If \(A\) and \(B\) are the points of intersection of the circle \(x^2+y^2-8 x=0\) and the hyperbola \(\frac{x^2}{9}-\frac{y^2}{4}=1\) and a point P moves on the line \(2 x-3 y+4=0\), then the centroid of \(\triangle \mathrm{PAB}\) lies on the line :
- A \(x+9 y=36\)
- B \(4 x-9 y=12\)
- C \(6 x-9 y=20\)
- D \(9 x-9 y=32\)
Answer & Solution
Correct Answer
(C) \(6 x-9 y=20\)
Step-by-step Solution
Detailed explanation
C: \(x^2+y^2-8 x=0\) \(\mathrm{H}: \frac{x^2}{9}-\frac{y^2}{4}=1\) By solving \(\frac{x^2}{9}-\left(\frac{8 x-x^2}{4}\right)=1\)…
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