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JEE MATHEMATICS MCQ: PARABOLA

1 Easy

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Application +4 | -1

The point of intersection of the curves whose parametric equations are

x=t^2+1, y=2 t

and

x=2 \mathrm{~s}, y=\frac{2}{s}

, is given by

View answer and explanation

Correct option: B

(2,2)

Option B ((2,2)) is correct because it is the point that satisfies both parametric equations when converted to Cartesian form.

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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