AP EAMCET · Maths · Ellipse
If the tangents drawn from a point \(P\) to the ellipse \(4 x^2+9 y^2-16 x+54 y+61=0\) are perpendicular, then the locus of \(P\) is
- A \(x^2+y^2-4 x+6 y+4=0\)
- B \(x^2+y^2-4 x+6 y=0\)
- C \(x^2+y^2-6 x+4 y+9=0\)
- D \(x^2+y^2-6 x+4 y=0\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2-4 x+6 y=0\)
Step-by-step Solution
Detailed explanation
Given ellipse: \(4 x^2+9 y^2-16 x+54 y+61=0\) \(4(x^2-4x) + 9(y^2+6y) + 61 = 0\) \(4((x-2)^2-4) + 9((y+3)^2-9) + 61 = 0\) \(4(x-2)^2 - 16 + 9(y+3)^2 - 81 + 61 = 0\) \(4(x-2)^2 + 9(y+3)^2 = 36\) \(\frac{(x-2)^2}{9} + \frac{(y+3)^2}{4} = 1\) Center \((h,k) = (2,-3)\), \(a^2=9\),…
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