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Question

The focus of the parabola is (1,1) and the tangent at the vertex has the equation x+y=1. Then:

A
equation of the parabola is (xy)2=2(x+y1)
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B
equation of the parabola is (xy)2=4(x+y1)
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C
the coordinates of the vertex are (12,12)
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D
length of the latus rectum is 22
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Solution

The correct option is B equation of the parabola is (xy)2=4(x+y1)
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,

xy+λ=0

This will pass through (1,1) if,

11+λ=0λ=0

So, the equation of the axis is xy=0

The vertex A is the point of intersection of xy=0 and x+y=1.

Solving these two equations, we get

x=12 and y=12

So, the coordinates of the vertex A are (12,12).

let (x1,y1) be the co-ordinates of k. Then,

x1+12=12,y1+12=12

x1=0, y1=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.
Equation of the directrix is

y0=1(x0), 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.

(x1)2+(y1)2=x+y2

2x2+2y24x4y+4=x2+y2+2xy

x2+y22xy4x4y+4=0

(xy)2=4x+4y4

(xy)2=4(x+y1)


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