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

Let a line perpendicular to the line \(2 x-y=10\) touch the parabola \(y^2=4(x-9)\) at the point \(P\). The distance of the point \(\mathrm{P}\) from the centre of the circle \(x^2+y^2-14 x-8 y+56=0\) is ...........

  1. A \(10\)
  2. B \(56\)
  3. C \(36\)
  4. D \(34\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(10\)

Step-by-step Solution

Detailed explanation

\( y^2=4(x-9) \) slope of tangent \(=\frac{-1}{2}\) Point of contact \(\mathrm{P}\left(9+\frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{\frac{-1}{2}}\right)\) \( P(13,-4) \) \( \text { center of circle } C(7,4) \) \( \text { distance } C P=\sqrt{(13-7)^2+(-4-4)^2} \)…