TS EAMCET · Maths · Hyperbola
If the line \(2 x+\sqrt{6} y=2\) touches the hyperbola \(x^2-2 y^3=4\), then the coordinates of the point of contact are
- A \(\left(\frac{1}{2}, \frac{1}{\sqrt{6}}\right)\)
- B \((4,-\sqrt{6})\)
- C \((4, \sqrt{6})\)
- D \((-2, \sqrt{6})\)
Answer & Solution
Correct Answer
(B) \((4,-\sqrt{6})\)
Step-by-step Solution
Detailed explanation
\(2 x+\sqrt{6} y=2\) \(\Rightarrow x^2-2 y^2=4\) Let point of contact \(=\left(x_1, y_1\right)\) Equation of tangent at \(\left(x_1, y_1\right)\) \(=x x_1-2 y y_1-4=0\) \(\because\) It represents same line as \(2 x+\sqrt{6} y-2=0\)…
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