AP EAMCET · Maths · Hyperbola
If the normal drawn to the hyperbola \(x y=16\) at \((8,2)\) meets the hyperbola again at a point \((\alpha, \beta)\), then \(|\beta|+\frac{1}{|\alpha|}=\)
- A \(40\)
- B \(34\)
- C \(28\)
- D \(54\)
Answer & Solution
Correct Answer
(B) \(34\)
Step-by-step Solution
Detailed explanation
\(\frac{dy}{dx} = -\frac{16}{x^2}\) \(m_{tangent} = -\frac{16}{8^2} = -\frac{1}{4}\) \(m_{normal} = 4\) \(y - 2 = 4(x - 8) \Rightarrow y = 4x - 30\) \(x(4x - 30) = 16 \Rightarrow 4x^2 - 30x - 16 = 0 \Rightarrow 2x^2 - 15x - 8 = 0\) \((x-8)(2x+1) = 0\) \(\alpha = -\frac{1}{2}\)…
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