the tangent to a parabola are `x-y=0 and x+y=0` If the focus of the parabola is F(2,3) then the equation of tangent at vertex is

Options

A. 4x-6y+5=0
B. 4x-6y+3=0
C. 4x-6y+1=0
D. 4x-6y+3/2=0

Answer

A

Explanation

(1)

We know that foot of perpendicular from the focus upon a tangent lies on the tangent at the vertex of the parabola.
Now, if the foot of perpendicular of (2,3) on the line x-y=0 is
`(x_(1),y_(1))` , then
`(x_(1)-2)/(1)=(y_(1)-3)/(-1)=(-(2-3))/(2)`
`orx_(1)=(5)/(2)andy_(1)=(5)/(2)`
If the foot of perpendicular of (2,3) on the line x+y=0 is `(x_(2),y_(2))`, then
`(x_(2)-2)/(1)=(y_(2)-3)/(1)=-(2+3)/(2)`
`orx_(2)=-(1)/(2)andy_(2)=(1)/(2)`
Now, the tangent at the vertex passes through the points `(5//2,5//2)and(-1//2,1//2)`. Then, its equation is
`y-(1)/(2)=(2)/(3)(x+(1)/(2))`
`or4x-6y+5=0`
The length of latus rectum of the parabola is `4xx` (Distance of locus from tangent at vertex)
`=4xx|(8-18+5)/(sqrt(52))|=(10)/(sqrt(13))`
Also, the distance between the focus and the tangent at vertex is `5//sqrt(13)`
We know that
`(1)/(SP)+(1)/(SQ)=(1)/(a)`
where a is `1//4`th of the length of latus rectum. Therefore,
`(1)/(SP)+(1)/(SQ)=(2sqrt(13))/(5)`