Question

The equation of the parabola whose focus is the point (0, 0) and the tangent at the vertex is x-y+1=0 is

• A. x²+y²-2xy-4x+4y-4=0
• B. x²+y²-2xy+4x–4y-4=0
• C. x²+y²+2xy-4x+4y-4=0
• D. x²+y²+2xy-4x–4y+4=0

Answer 1

Answer: (C)

x²+y²+2xy-4x+4y-4=0

Answer 2

Let focus is S(0, 0) and A is the vertex of parabola. Take any point Z such that AS=AZ. Given tangent at vertex is

x-y+1 = 0.

Since directrix is parallel to the tangent at the vertex.

∴ Equation of directrix is x – y + λ = 0, where λ is constant.

∴ A is midpoint of SZ, ∴ SZ = 2.SA

⇒ $\frac{|0 - 0 + \lambda|}{\sqrt{1^{2} + ( - 1)^{2}}}\mspace{6mu} = 2x\frac{|0 - 0 + 1|}{\sqrt{1^{2} + ( - 1)^{2}}} \Rightarrow \mspace{6mu}|\lambda| = 2$ ie., λ = 2.

$\because$ Directrix in this case always lies in II^nd quadrant.

∴ λ = 2

Hence equation of directrix is x – y + 2 = 0.

Now, P be any point on parabola.

∴ SP = PM ⇒ SP² = PM²

⇒ x² + y² + 2xy – 4x + 4y – 4 = 0.