The equation of the line of the shortest distance between the parabola $y^{2} = 4 x$ and the circle $x^{2} + y^{2} - 4 x - 2 y + 4 = 0$

x+y=3

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x-y=3

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2x+y=5

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done

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Solution

The correct option is A x+y=3
Line of shortest distance is normal for both parabola and circle
Centre of circle is (2, 1)
Equation of normal to circle is $y - 1 = m (x - 2) \Rightarrow y = m x + (1 - 2 m) (1)$
Equation of normal for a parabola is $y = m x - 2 a m - a m^{3} (3)$
Comparing (1) and (2)
$a m^{3} = - 1 \Rightarrow m^{3} = - 1 \Rightarrow m = - 1 (a = 1)$
Equation is $y - 1 = x + 2 \Rightarrow x + y = 3$

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Similar questions

y^{2} = 4 x

x^{2} + y^{2} - 4 x - 2 y + 4 = 0

y^{2} = 4 x

x^{2} + y^{2} - 4 x - 2 y + 4 = 0

y^{2} = 4 x

x^{2} + y^{2} + 6 x - 12 y + 20 = 0

x^{2} + y^{2} - 2 x - 6 y + 6 = 0

(2, 1)

x^{2} + y^{2} - 4 x - 6 y - 3 = 0

x^{2} + y^{2} + 4 x - 2 y - 4 = 0

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