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JEE MATHEMATICS MCQ: PARABOLA

1 Easy

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Application +4 | -1

Equation of the common tangent to

y^2=4 x

and

x^2=32 y

is

View answer and explanation

Correct option: A

x+2 y+4=0

Option A (

x+2y+4=0

) is correct because it represents the common tangent found by solving the equations of tangents to both curves and equating them.

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The equation of a common tangent to the parabolas

y=x^2

y=-(x-2)^2

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