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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(P\) be the point on the parabola \({y^2} = 8x\) which is at a minimun distance from the centre \(C\) of the circle \({x^2} + {\left( {y + 6} \right)^2} = 1\) . Then the equation of the circle , passing through \(C\) and having its centre at \(P \) is :
- A \({x^2} + {y^2} - \frac{x}{4} + 2y - 24 = 0\)
- B \(\;{x^2} + {y^2} - 4x + 9y + 18 = 0\)
- C \(\;{x^2} + {y^2} - 4x + 8y + 12 = 0\)
- D \(\;{x^2} + {y^2} - x + 4y - 12 = 0\)
Answer & Solution
Correct Answer
(C) \(\;{x^2} + {y^2} - 4x + 8y + 12 = 0\)
Step-by-step Solution
Detailed explanation
Minimum distance occurs along common normal Let normal to parabola be \(y+t x=2.2 . t+2 t^{3}\) pass through \((0,-6):\) \(-6=4 t+2 t^{3} \Rightarrow t^{3}+2 t+3=0\) \(\Rightarrow \mathrm{t}=-1\) (only real value) \(\therefore P(2,-4)\)…
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