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the axis of the parabola is the line y=x and the distance of vertex from origin is √2 and that of origin from its focus is 2√2 . If vertex and focus both lie in 1 quadrant, then the equation of parabola is
First let's find the equation of a horizontal parabola in set of coordinate axes rotated 45° from the original set. The distance from the origin to the vertex is 2. The vertex is: (h,k) = (2,0) The directed distance from the vertex to the focus is p. p = 2√2 The equation of the parabola is: 4p(x' - 2) = (y' - 0)² 4(2√2)(x' - 2) = y'² 8√2(x' - 2) = y'² ________ x' = xcosθ + ysinθ y' = -xsinθ + ycosθ x' = xcos(π/4) + ysin(π/4) = x/√2 + y/√2 y' = -xsinθ + ycosθ = -x/√2 + y/√2 Now plug the values into the equation to get the equation in the original set of coordinates. 8√2(x' - 2) = y'² 8√2[(x/√2 + y/√2) - 2] = (-x/√2 + y/√2)² 8x + 8y - 16√2 = x²/2 - xy + y²/2 16x + 16y - 32√2 = x² - 2xy + y² x² - 2xy + y² - 16x - 16y + 32√2 = 0
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