Prove that the equation of the parabola whose focus is (0, 0) and tangent at the vertex is `x-y+1 = 0 ` is `x^2 + y^2 + 2xy - 4x + 4y - 4=0`.

Options

A. `x^(2)+y^(2)-2xy-4x+4y-4=0`
B. `x^(2)+y^(2)-2xy+4x-4y-4=0`
C. `x^(2)+y^(2)+2xy-4x+4y-4=0`
D. `x^(2)+y^(2)+2xy-4x-4y+4=0`

Answer

C

Explanation

We know that if P(x, y) is a point on a parabola and PM and PN are perpendiculars drawn from P on the axis and tangent at the vertex of the parabola, Then, the equation of the parabola is given by
`PM^(2)=("Latusrectum ") PN`
It is given that x-y+1=0 is the equation of the tangent at the vertex. Axis is perpendicular to x-y+1=0 and passes through the focus. So, its equation is x + y = 0.
Also,
`"Latusrectum = 4 (AS)"=4(|0-0+1|)/(sqrt(1+1))=2sqrt2`
So, the equation of the parabola is given by
`PM^(2)=2sqrt2(PN)`

`rArr" "|(x+y)/sqrt2|^(2)=2sqrt2(|x-y+1|)/sqrt2`
`rArr" "(x+y)^(2)=4(x-y+1)rArrx^(2)+y^(2)+2xy-4y-4=0`