AP EAMCET · Maths · Hyperbola
The equation of the transverse axis of hyperbola \((x-3)^2+(y+1)^2=(4 x+3 y)^2\) is
- A \(3 x+4 y=13\)
- B \(3 x-4 y=13\)
- C \(4 x-3 y=13\)
- D \(3 x-4 y=9\)
Answer & Solution
Correct Answer
(B) \(3 x-4 y=13\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} (x-3)^2+(y+1)^2 & =(4 x+3 y)^2 \\ (x-3)^2+(y+1)^2 & =25\left(\frac{4 x+3 y}{5}\right)^2 \\ (x-3)^2+(y+1)^2 & =25\left(\frac{4 x+3 y}{25}\right)^2 \\ \sqrt{(x-3)^2+(y+1)^2} & =5\left(\frac{4 x+3 y}{\sqrt{5}}\right)^2 \end{aligned}\) \(\therefore\) It is of the…
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