JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the common tangents to the parabola, \(x^2 = 4y\) and the circle, \(x^2 + y^2 = 4\) intersect at the point \(P\), then find the square of the slope of the line:
- A \(\sqrt 2 + 1\)
- B \(2\left( {3 + 2\sqrt 2 } \right)\)
- C \(2\left( {\sqrt 2 + 1} \right)\)
- D \(3 + 2\sqrt 2 \)
Answer & Solution
Correct Answer
(C) \(2\left( {\sqrt 2 + 1} \right)\)
Step-by-step Solution
Detailed explanation
Tangent to \({x^2} + {y^2} = 4\) \(y = mx \pm 2\sqrt {1 + {m^2}} \) Also, \({x^2} = 4y\) \( \Rightarrow {x^2} = 4mx + 8\sqrt {1 + {m^2}} \) or \({x^2} = 4mx - 8\sqrt {1 + {m^2}} \) For \(D=0\) we have; \(16{m^2} + 4.8\sqrt {1 + {m^2}} = 0\)…
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