Consider the parabola whose focus is at (0,0) and tangent at vertex is x-y+1=0
The length of latus rectum is

Options

A. `4sqrt(2)`
B. `2sqrt(2)`
C. `8sqrt(2)`
D. `3sqrt(2)`

Answer

B

Explanation

(2)

The distance between the focus and tangent at the vertex is
`(|0-0+1|)/(sqrt(1^(2)+1^(2)))=(1)/(sqrt(2))`
The directrix is the line parallel to the tangent vertex ant at a distance `2xx1sqrt(2)` from the focus.
Let the equation of the directrix be
`x-y+lamda=0`
where `(lamda)/(sqrt(1^(2)+1^(2)))=(2)/(sqrt(2))`
`:." "lamda=2`
Let P (x,y) be any moving point on the parabola. Then, OP=PM
`orx^(2)+y^(2)=((x-y+2)/(sqrt(1^(2)+1)))^(2)`
`or2x^(2)+=2y^(2)=(x-y+2)^(2)`
`orx^(2)+y^(2)+2xy-4x+4y-4=0`
Latus rectum length `=2xx` (Distance of focus from directrix)
`2|(0-0+2)/(sqrt(1^(2)+1^(2)))|=2sqrt(2)`
Solving the parabola with the x-axis, we get
`x^(2)-4x-4=0`
`orx=(4pmsqrt(32))/(2)=2pm2sqrt(2)`
Therefore, the length of chord on the x-axis is `4sqrt(2)`.
Since the chord 3x+2y=0 passes through the focus, it is focus chord.
Hence, tangents at the extremities of chord are perpendicular.