JEE MATHEMATICS MCQ: PARABOLA

1 Easy

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Application +4 | -1

The equation of the chord of

y^2=8 x

which is bisected at

(2,-3)

, is

View answer and explanation

Correct option: A

4 x+3 y+1=0

Option A (

4 x+3 y+1=0

) is correct because the equation of the chord bisected at a given point is found using the formula T = S', where T and S' are calculated based on the given parabola and the point of bisection.

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