JEE MATHEMATICS MCQ: PARABOLA

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The possible coordinate of the point on

y^2=8 x

, which is closest from

\mathrm{x}^2+(\mathrm{y}+6)^2=1

is :

View answer and explanation

Correct option: A

(2,-4)

Option A ((2,-4)) is correct because it represents the point on the parabola closest to the given circle, found by minimizing the distance between the point on the parabola and the center of the circle.

More Questions from PARABOLA

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The equation of a common tangent to the parabolas

y=x^2

and

y=-(x-2)^2

is A point

P

moves such that slope of one tangent drawn from it to parabola

y^2=16 x

is four times the slope of other tangent, then the length of latus rectum of the conic represented by locus of

P

, is Let

P,Q,R

be three points on parabola

{y^2} = 4x

and normal at

P

and

R

meet at

Q

, then the locus of the mid-point of the chord

PR

is a parabola whose vertex is at The tangents at the end points of any chord through

(1,0)

to the parabola

y^2+4 x=8

intersect The focus of the parabola

x^2=-16 y

is If

I

denotes the semi-latusrectum of the parabola

{y^2} = 4ax

and

SP

and

SQ

denote the segments of any focal chord

PQ,S

being the focus, then

SP,l

and

SQ

are in the relation