(i) y² = 10 x 

A. Compare this equation with ( y - k )² = 4a( x - h ) 

then we get k = 0 , h = 0 . So vertex = ( h , k ) = ( 0 , 0 )

Also 4a = 10 ( as per given data ) . So a = 10 / 4 = 5 / 2

As we know that focus of the parabola is S( a, 0 )  [since a > 0]. Then coordinates of S = ( 5 / 2 , 0 )

Since the given parabola is symmetric about x axis and S lies on x axis 

Equation of directrix is x + a = 0 , i.e  x + 5 / 2 = 0 , 2x + 5 = 0

we know that length of latus rectum is 4a . So length = 4( 5 / 2 ) = 10

(ii) x² = 6y

A. Compare the given equation with ( x - h )² = 4a( y - k )

then we get h = 0 , k = 0 .So vertex = ( h , k ) = ( 0 , 0 )

Also 4a = 6 ( as per given data ) . So a = 6 / 4 = 3 / 2

As  we know that focus of this parabola is ( 0 , a )   [ since a > 0 ]

Then S ( 0 , 3 / 2 )

Since the given parabola is symmetric about y axis and S lies on y axis 

Equation of directrix = y + a = 0 , i.e y + 3 / 2  = 0 , 2y + 3 = 0

we know that length of the latus rectum = 4a . So length = 4( 3 / 2 ) = 6