JEE MATHEMATICS MCQ: PARABOLA

1 Easy

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Analyzing +4 | -1

Consider the parabola with vertex

\left( {\frac{1}{2},\frac{3}{4}} \right)

and the directrix

y = \frac{1}{2}.

Let

P

be the point where the parabola meets the line

x = - \frac{1}{2}.

If the normal to the parabola at

P

intersects the parabola again at the point

Q

then

{(PQ)^2}

is equal to

View answer and explanation

Correct option: B

\frac{{125}}{{16}}

Option B (

\frac{{125}}{{16}}

) is correct because it is the square of the distance between the point P on the parabola and the point Q where the normal at P intersects the parabola again, calculated using the parabola's equation and normal properties.

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