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JEE MATHEMATICS MCQ: PARABOLA

1 Easy

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Application +4 | -1

The equation of a common tangent to the parabolas

y=x^2

and

y=-(x-2)^2

is

View answer and explanation

Correct option: B

y=4(x-1)

Option B is correct because the common tangent is found by equating the tangent equations of both parabolas and solving for the slope.

More Questions from PARABOLA

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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

P,Q,R

be three points on parabola

{y^2} = 4x

P

R

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

I

denotes the semi-latusrectum of the parabola

{y^2} = 4ax

SP

SQ

denote the segments of any focal chord

PQ,S

being the focus, then

SP,l

SQ

are in the relation The equation of the parabola with its vertex at the origin, axis on the

y - axis

and passing through the point

\left( {6, - 3} \right)