The point of contact of the line 2 x+ 6 y=2 and the hyperbola x^ 2 -2 y^ 2 =4 is,
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
The point of contact of the line \(2 x+\sqrt{6} y=2\) and the hyperbola \(x^{2}-2 y^{2}=4\) is,
- \((4,-\sqrt{6})\)
- \([\sqrt{6}, 1]\)
- \(\left(\frac{1}{2}, \frac{1}{\sqrt{6}}\right)\)
- \(\left(\frac{1}{6}, \frac{3}{2}\right)\)
Answer
(A) (4,- 6 )
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