find the equation of parabola whose focus is (1,1) and tangent at the vertex is x + y = 1 - Maths - Principle of Mathematical Induction

Question

find the equation of parabola whose focus is (1,1) and tangent
at the vertex is x + y = 1

Ajanta Trivedi · Accepted Answer

let S be the focus and A be the vertex of the parabola. let K be the point of intersection of the axis and directrix.

since axis is a line passing through S(1,1) and perpendicular to x+y=1.

so, let the equation of the axis be x-y+λ=0

this will pass through (1,1) if 1-1+λ=0 ⇒λ=0

so the equation of the axis is x-y=0

the vertex A is the point of intersection of x-y=0 and x+y=1. solving these two equations, we get x=1/2 , y=1/2

so, the coordinates of the vertex A are (1/2,1/2).

let $(x_{1}, y_{1})$ be the coordinates of K. then,

$\frac{x_{1} + 1}{2} = \frac{1}{2}, \frac{y_{1} + 1}{2} = \frac{1}{2} \Rightarrow x_{1} = 0, y_{1} = 0$

so, the co-ordinates of K are (0,0).

since directrix is a line passing through K(0,0) and parallel to x+y=1.

therefore, equation of the directrix is y-0=-1(x-0), i.e. x+y=0

let P(x,y) be any point on the parabola. then,

distance of P from the focus S= distance of P from the directrix

$\sqrt{(x - 1)^{2} + (y - 1)^{2}} = \frac{x + y}{\sqrt{2}} 2 x^{2} + 2 y^{2} - 4 x - 4 y + 4 = x^{2} + y^{2} + 2 x y x^{2} + y^{2} - 2 x y - 4 x - 4 y + 4 = 0$