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Options

A. vertex is `(2a//3,0)`
B. directrix is x=0
C. latus rectum is `2a//3`
D. focus is (a,0)

Answer

A, D

Explanation

1,4

Tangent at point `P(at,2at)` is `ty=x+at^(2)`.
It meets the x-axis at `(-at^(2),0)`.
Normal at point P is `y=-tx+2at+at^(3)`.
It meets the x-axis at `(2a+at^(2),0)`.
Let the centroid of triangle PNT be `G-=(h,k)`. Then,
`h=(2a+at^(2))/(a)andk=(2at)/(3)`
Eliminating t, we get
`:." "((3h-2a)/(a))=(9k^(2))/(4a^(2))`
So, the required parabola is
`(9y^(2))/(4a^(2))=((3x-2a))/(a)=(3)/(a)(x-(2a)/(3))`
`ory^(2)=(4a)/(3)(x-(2a)/(3))`
`"Vertex"-=((2a)/(3),0),"focus"-=((2a)/(3)+(a)/(3),0)-=(a,0)`