find the equation of parabola whose focus is (1,1) and tangent at the vertex is x + y = 1
Answers
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We have this equation:
x + y = 1
x + y = 1
y = - x + 1
The slope of the tangent here is -1
And the slope of the line perpendicular to the tangent is 1
The line will pass through the focus y = x + c which will pass through (1,1)
The equation for this would be: y = x
The point of intersection of x + y = 1 and x = y can be obtained by:
y + y = 1
2y = 1
y = 1/2
The vertex for this is : (1/2, 1/2)
The distance between the focus and vertex is √[2x(1/2)²] = √2/2
The equation of parabola would be (y -1/2)² = (2x√2) x (x-1/2) y² + 1/4 - y = 2√2*x - √2
= 4y² - 4y - 8 x √2 *x + 4*√2 + 1 = 0
If there is any confusion please leave a comment below.
Answer: The equation of parabola is 4y^2-4y-8(2^(1/2))x+4(2^(1/2))+1=0
Step-by-step explanation:
Given that the equation of tangent at vertex, x+y=1
x+y=1 => y=-x+1
Slope of the tangent is -1. So slope of the axis of the parabola which is perpendicular to the tangent would be 1.
The equation of axis of the parabola would be y= x+c
We know the focus of the parabola passes through the axis. So (1,1) passes through y=x+c
This gives the equation of the axis as y=x
The intersection point of x+y=1 and x=y gives the vertex.
So, the intersection of x+y=1 and x=y would be (1/2,1/2)
The vertex is (1/2,1/2).
Now, we find for ‘a’.
a= distance between focus and vertex
a= ((1-1/2)^2+(1-1/2)^2)^(1/2)= ((1/2)^2+(1/2)^2)^(1/2)= 2^(1/2)/2
The general equation of parabola is given by
(y-h)^2= 4a(x-k) where (h,k) is focus.
Now, put all the values in the general equation of parabola as follow:
(y-1/2)^2= 4x2^(1/2)/2(x-1/2)
Solving the above equation, we get:
4y^2-4y-8(2^(1/2))x+4(2^(1/2))+1=0