JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the shortest distance from \((\mathrm{a}, 0), \mathrm{a}\gt0\), to the parabola \(y^2=4 x\) be 4 . Then the equation of the circle passing through the point \((a, 0)\) and the focus of the parabola, and having its centre on the axis of the parabola is :
- A \(x^2+y^2-10 x+9=0\)
- B \(x^2+y^2-6 x+5=0\)
- C \(x^2+y^2-4 x+3=0\)
- D \(x^2+y^2-8 x+7=0\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2-6 x+5=0\)
Step-by-step Solution
Detailed explanation
Normal at P \(\begin{aligned} & y+ t x=2 t+t^3 \\ & \uparrow \\ &(\mathrm{a}, 0) \\ & \mathrm{at}= 2 \mathrm{t}+\mathrm{t}^3 \\ & \mathrm{a}= 2+\mathrm{t}^2 \\ & \mathbb{R}\left(2+\mathrm{t}^2, 0\right) \end{aligned}\)…
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