The axis of a parabola is along the line y=x. | vedclass

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(A) The axis of the parabola is $y=x$. The vertex $A$ and focus $S$ lie on this line in the first quadrant. Since the distance of $A$ from $(0,0)$ is

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$x=(t+1)^2, y=(t-1)^2$

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(A) The axis of the parabola is $y=x$. The vertex $A$ and focus $S$ lie on this line in the first quadrant. Since the distance of $A$ from $(0,0)$ is $\sqrt{2}$,the coordinates of $A$ are $(1,1)$. Since the distance of $S$ from $(0,0)$ is $2\sqrt{2}$,the coordinates of $S$ are $(2,2)$. The distance $AS = a = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$. The directrix is perpendicular to the axis $y=x$ and passes through a point $Z$ such that $A$ is the midpoint of $SZ$. Since $S=(2,2)$ and $A=(1,1)$,$Z=(0,0)$. The equation of the directrix is $x+y=0$. By the definition of a parabola,the distance from any point $(x,y)$ to the focus $(2,2)$ equals the distance to the directrix $x+y=0$: $(x-2)^2 + (y-2)^2 = \frac{(x+y)^2}{2}$. $2(x^2 - 4x + 4 + y^2 - 4y + 4) = x^2 + y^2 + 2xy$. $x^2 + y^2 - 2xy - 8x - 8y + 16 = 0$,which is $(x-y)^2 = 8(x+y-2)$. Substituting $x=(t+1)^2$ and $y=(t-1)^2$: $((t+1)^2 - (t-1)^2)^2 = (4t)^2 = 16t^2$. $8((t+1)^2 + (t-1)^2 - 2) = 8(t^2 + 2t + 1 + t^2 - 2t + 1 - 2) = 8(2t^2) = 16t^2$. Both sides match,so the parametric form is $x=(t+1)^2, y=(t-1)^2$.

The axis of a parabola is along the line $y=x$. The distance of its vertex $A$ from $(0,0)$ is $\sqrt{2}$ and that of its focus $S$ from $(0,0)$ is $2\sqrt{2}$. If $A$ and $S$ lie in the first quadrant,then the equation of the parabola in parametric form is

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