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 ...........
- A \(10\)
- B \(56\)
- C \(36\)
- D \(34\)
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 \(\mathrm{P}\left(9+\frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{\frac{-1}{2}}\right)\) \( P(13,-4) \) \( \text { center of circle } C(7,4) \) \( \text { distance } C P=\sqrt{(13-7)^2+(-4-4)^2} \)…
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