TS EAMCET · Maths · Hyperbola
The slope of the tangent drawn from the point \((1,1)\) to the hyperbola \(2 x^2-y^2=4\) is
- A 2
- B \(\frac{-2 \pm \sqrt{6}}{2}\)
- C \(-1 \pm \sqrt{6}\)
- D \(\frac{-2 \pm \sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(C) \(-1 \pm \sqrt{6}\)
Step-by-step Solution
Detailed explanation
We know that equation of tangents to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) are \(y=m x \pm \sqrt{a^2 m^2-b^2}\) Here, equation of hyperbola is \(2 x^2-y^2=4 \Rightarrow \frac{x^2}{2}-\frac{y^2}{4}=1\) \(\therefore\) Equation of tangents are…
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