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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Parabola

Let a line perpendicular to the line 2 x-y=10 touch the parabola...

Let a line perpendicular to the line 2x−y=102 x-y=10 touch the parabola y2=4(x−9)y^2=4(x-9) at the point PP. The distance of the point PP from the centre of the circle x2+y2−14x−8y+56=0x^2+y^2-14 x-8 y+56=0 is __________

Correct answer

0

Step-by-step solution

y2=4(x−9)slopeofy^2=4(x-9)slope oftangent =−12Point\frac{-1}{2} Point of contact P(9+1(−12)2,2×1−12)\mathrm{P}\left(9+\frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{\frac{-1}{2}}\right)P(13,−4) center of circle C(7,4) distance CP=(13−7)2+(−4−4)2=10\begin{aligned} & P(13,-4) \\ & \text{ center of circle } C(7,4) \\ &\text{ distance } C P=\sqrt{(13-7)^2+(-4-4)^2} \\ & =10\end{aligned}$$
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