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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

माना परवलय \(x^{2}=4 y\) पर \(P\) एक बिंदु है। यदि \(P\) की वृत्त \(x^{2}+y^{2}+6 x+8=0\) के केन्द्र से दूरी न्यूनतम है, तो परवलय के बिंदु \(P\) पर स्पर्शरेखा का समीकरण है

  1. A \(x+4y - 2 = 0\)
  2. B \(x+ 2y = 0\)
  3. C \(x+y+ 1 = 0\)
  4. D \(x -y+3=0\)
Verified Solution

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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