Question

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

Answer 1

Thank you for asking this question. Here is your answer:

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 2

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