JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(P\) be a point on the parabola, \(x^2 = 4y.\) If the distance of \(P\) from the centre of the circle, \(x^2 + y^2 + 6x + 8 = 0\) is minimum, then the equation of the tangent to the parabola at \(P,\) is
- A \(x+4y - 2 = 0\)
- B \(x+ 2y = 0\)
- C \(x+y+ 1 = 0\)
- D \(x -y+3=0\)
Answer & Solution
Correct Answer
(C) \(x+y+ 1 = 0\)
Step-by-step Solution
Detailed explanation
Let \(\frac{{{t^2} - 0}}{{2t + 3}}\) be any point on the parabola. Centre of the given circle \(C = \left( { - g, - f} \right) = \left( { - 3,0} \right)\) For \(PC\) to be minimum, it must be the normal to the parabola at \(P\). Slop of line…
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