JEE MATHEMATICS MCQ: PARABOLA

1 Medium

1 of 1

Application +4 | -1

The locus of the point of intersection of tangents to the parabola

{y^2} = 4\left( {x + 1} \right)

and

{y^2} = 8\left( {x + 2} \right)

which are perpendicular to each other is

View answer and explanation

Correct option: C

x + 3 = 0

The correct answer is C (

x + 3 = 0

) because the locus of the intersection of perpendicular tangents to parabolas is the directrix, and the directrix of the combined parabola system is x = -3.

More Questions from PARABOLA

Practice All Chapter MCQs →

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