4 x which is at the shortest distance from the centre of the circle x 2- y 2-4 x -16 y -64-0 is P - - then A point on the parabola y 2- -2-2-A- 8B- 16C- 32D- 64

The correct option is C 32C = (2, 8), CP will be common normal ⇒ y = mx 2m m^3 is P.T. (2, 8) ⇒ m = 2 ⇒ P = (4, 4) ∴α ^2 + β ^2 = 32. ...

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Standard XII

Mathematics

Double Ordinate

4 x which is ...

Question

A point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the centre of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ is $P (α, β)$ then $α^{2} + β^{2} =$ .

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Solution

The correct option is C 32
C = (2, 8), CP will be common normal
$\Rightarrow y = m x - 2 m - m^{3}$ is P.T. (2, 8)
$\Rightarrow m = - 2 \Rightarrow$ P = (4, 4)
$∴ α^{2} + β^{2} = 32$ .

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Similar questions

Q. Let

P

be the point on the parabola

y^{2} = 4 x

which is at the shortest distance from the centre S of the circle

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

. Let

Q

be the point on the circle dividing the line segment

S P

internally. Then,

Q. Let

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y^{2} = 4 x

which is at the shortest distance from the centre

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x^{2} + y^{2} - 4 x - 16 y + 64 = 0.

Let

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be the point on the circle dividing the line segment

S P

internally. Then

Q. Let

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which is at the shortest distance from the centre

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x^{2} + y^{2} - 4 x - 16 y + 64 = 0.

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S P

internally. Then

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x^{2} + y^{2} = 5

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then the value of

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A point on the parabola y2=4x which is at the shortest distance from the centre of the circle x2+y2−4x−16y+64=0 is P(α,β) then α2+β2=.

The correct option is C 32C = (2, 8), CP will be common normal ⇒y=mx−2m−m3 is P.T. (2, 8) ⇒m=−2⇒ P = (4, 4) ∴α2+β2=32.

A point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the centre of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ is $P (α, β)$ then $α^{2} + β^{2} =$ .

The correct option is C 32
C = (2, 8), CP will be common normal
$\Rightarrow y = m x - 2 m - m^{3}$ is P.T. (2, 8)
$\Rightarrow m = - 2 \Rightarrow$ P = (4, 4)
$∴ α^{2} + β^{2} = 32$ .