JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the shortest distance of the parabola \(y^2=4 x\) from document security settings le \(x^2+y^2-4 x-16 y+64=0\) is \(\mathrm{d}\), then \(\mathrm{d}^2\) is equal to :
- A \(16\)
- B \(24\)
- C \(20\)
- D \(36\)
Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
Equation of normal to parabola \(y=m x-2 m-m^3\) this normal passing through center of circle \((2,8)\) \( 8=2 \mathrm{~m}-2 \mathrm{~m}-\mathrm{m}^3\) \( \mathrm{~m}=-2\) So point \(P\) on parabola \(\Rightarrow\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(4,4)\) And…
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