AP EAMCET · Maths · Hyperbola
The locus of a point which is at a distance of 2 units from the line \(2 x-3 y+4=0\) and at a distance of \(\sqrt{13}\) units from a point \((5,0)\) is
- A \(8 x^2+12 x y+56 x-24 y+84=0\)
- B \(12 x y-5 y^2-56 x+24 y+84=0\)
- C \(8 \mathrm{x}^2+12 \mathrm{xy}+\mathrm{y}^2-56 \mathrm{x}+24 \mathrm{y}+84=0\)
- D \(8 x^2+12 x y-7 y^2-56 x+24 y+84=0\)
Answer & Solution
Correct Answer
(B) \(12 x y-5 y^2-56 x+24 y+84=0\)
Step-by-step Solution
Detailed explanation
Let \(P(h, k)\) be the point whose locus is to be find. given \(\sqrt{(h-5)+(k-0)^2}=\sqrt{13}\) \[ \Rightarrow h^2=-k^2+10 h-12...(1) \] also given, \(\frac{2 h-3 k+4}{\sqrt{4+9}}=2\) Squaring both sides, we get- \[ \Rightarrow 4 h^2+9 k^2+16-12 h k-24 k+16 h=52 \] Putting…
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