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

Expert Answers

We have the equation of the tangent at the vertex equal to x + y = 1.

x + y = 1

=> y = -x + 1

The slope of  the tangent is -1, the slope of the line perpendicular to the tangent is 1. This line passes through...

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We have the equation of the tangent at the vertex equal to x + y = 1.

x + y = 1

=> y = -x + 1

The slope of  the tangent is -1, the slope of the line perpendicular to the tangent is 1. This line passes through the focus. y = x + c passes through (1,1). This gives the equation of the line as y = x

The point of intersection of x + y = 1 and x = y can be got by

y + y = 1

=> 2y = 1

=> y = 1/2

The vertex is (1/2, 1/2)

The distance between the focus and vertex is sqrt [2*(1/2)^2] = sqrt 2 / 2

The equation of the parabola is (y - 1/2)^2 = (2*sqrt 2 ) * ( x - 1/2)y^2 + 1/4 - y = 2*sqrt 2*x - sqrt 2

=> 4y^2 - 4y - 8*sqrt 2*x + 4*sqrt 2 + 1 = 0

The required equation of the parabola is 4y^2 - 4y - 8*sqrt 2*x + 4*sqrt 2 + 1 = 0

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