The shortest distance between the line $x{\rm{ }}-{\rm{ }}y{\rm{ }} = {\rm{ }}1\,$ and the curve ${x^2} = 2y$ is

JEE MATHEMATICS MCQ: PARABOLA

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The shortest distance between the line

x\mathrm{ }-\mathrm{ }y\mathrm{ } = \mathrm{ }1\,

and the curve

{x^2} = 2y

View answer and explanation

Correct option: B

\frac{1}{{2\sqrt 2 }}

Option B (

\frac{1}{{2\sqrt 2 }}

) is correct because it is the shortest distance found by finding a tangent to the curve parallel to the line and then calculating the perpendicular distance.

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