JEE MATHEMATICS MCQ: PARABOLA

1 Easy

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Analyzing +4 | -1

The parabolas :

a x^2+2 b x+c y=0

and

d x^2+2 e x+f y=0

intersect on the line

y=1

.

If

a, b, c, d, e, f

are positive real numbers and

a, b

c

are in G.P., then

View answer and explanation

Correct option: A

\frac{d}{a}, \frac{e}{b}, \frac{f}{c}

are in A.P.

Option A (

\frac{{d}}{{a}}, \frac{{e}}{{b}}, \frac{{f}}{{c}}

are in A.P.) is correct because by finding the intersection point and using the G.P. condition, we can show that the ratios are in arithmetic progression.

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