Let P be the point on the parabola y² = 4x which is at the shortest distance from the center S of the circle x² + y² - 4x - 16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then- [JEE Adv 2016 P2]

the x-intercept of the normal to the parabola at P is 6

the slope of the tangent to the circle at Q is 1/2

Test Index

Parabola

Section: Mathematics

Question : 14 of 31

Marks: +1, -0

y^2 = 4x

x^2 + y^2 - 4x - 16y + 64 = 0

Solution:

y^2=4x

point P lies on normal to parabola passing through centre of circle

y+tx=2t+t^3

......(i)

8+2t = 2t = t^3

t=2

P(4,4)

SP = \sqrt{(4-2)^2 + (4-8)^2}

SP = 2\sqrt{5}

SQ=2

\Rightarrow PQ = 2\sqrt{5} - 2

\frac{SQ}{QP} = \frac{1}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{4}

To find x intercept

put put y = 0 in (i)

\Rightarrow x = 2 + t^2

x=6

∵Slope of common normal = – t = –2

∴ slope of tangent

=\frac{1}{2}

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