STATEMENT-1: The curve y = -frac{x^2}{2} + x | vedclass

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Question

(A) Given the equation of the parabola: $y = -\frac{1}{2}x^2 + x + 1$.To find the axis of symmetry,we complete the square:$y = -\frac{1}{2}(x^2 - 2x)

Vedclass · Accepted Answer

$Statement-1$ is True,$Statement-2$ is True; $Statement-2$ is a correct explanation for $Statement-1$

Vedclass · Answer

(A) Given the equation of the parabola: $y = -\frac{1}{2}x^2 + x + 1$.To find the axis of symmetry,we complete the square:$y = -\frac{1}{2}(x^2 - 2x) + 1$$y = -\frac{1}{2}(x^2 - 2x + 1 - 1) + 1$$y = -\frac{1}{2}(x - 1)^2 + \frac{1}{2} + 1$$y - \frac{3}{2} = -\frac{1}{2}(x - 1)^2$.This is in the form $(x - h)^2 = 4a(y - k)$,which represents a parabola symmetric about the line $x = h$. Here,$h = 1$,so the curve is symmetric about $x = 1$.$Statement-1$ is True. $Statement-2$ is a standard property of a parabola,which is also True. Since the axis of symmetry for this parabola is indeed $x = 1$,$Statement-2$ correctly explains $Statement-1$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

$STATEMENT-1$: The curve $y = -\frac{x^2}{2} + x + 1$ is symmetric with respect to the line $x = 1$. Because
$STATEMENT-2$: $A$ parabola is symmetric about its axis.

Similar Questions

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$PQ$ is any focal chord of the parabola $y^2=32x$. The length of $PQ$ can never be less than: ............ $unit$

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$STATEMENT-1$: The curve $y = -\frac{x^2}{2} + x + 1$ is symmetric with respect to the line $x = 1$. Because$STATEMENT-2$: $A$ parabola is symmetric about its axis.