The focus of the parabola $(y-2)^2=20(x+3)$ is

JEE MATHEMATICS MCQ: PARABOLA

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The focus of the parabola

(y-2)^2=20(x+3)

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Correct option: C

(2,2)

Option C (2,2) is correct because the focus of the parabola is found by comparing the given equation with the standard equation and calculating the focus using the vertex and the value of 'a'.

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