A point on the parabola whose focus is S(1,-1 | vedclass

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Question

(A) The axis of the parabola is the line passing through the focus $S(1,-1)$ and vertex $A(1,1)$. Since the $x$-coordinates are the same,the axis is t

Vedclass · Accepted Answer

$\left(3, \frac{1}{2}\right)$

Vedclass · Answer

(A) The axis of the parabola is the line passing through the focus $S(1,-1)$ and vertex $A(1,1)$. Since the $x$-coordinates are the same,the axis is the vertical line $x=1$.The distance between the vertex $A(1,1)$ and focus $S(1,-1)$ is $a = \sqrt{(1-1)^2 + (1 - (-1))^2} = 2$.Since the vertex is above the focus,the parabola opens downwards. The directrix is a horizontal line at a distance $a=2$ above the vertex $A(1,1)$.The equation of the directrix is $y = 1 + 2 = 3$.By the definition of a parabola,for any point $P(x,y)$ on it,the distance to the focus equals the distance to the directrix: $PS^2 = PM^2$.$(x-1)^2 + (y+1)^2 = (y-3)^2$.$(x-1)^2 + y^2 + 2y + 1 = y^2 - 6y + 9$.$(x-1)^2 = -8y + 8 = 8(1-y)$.Checking option $A$: For $\left(3, \frac{1}{2}\right)$,$(3-1)^2 = 2^2 = 4$ and $8(1 - \frac{1}{2}) = 8(\frac{1}{2}) = 4$.Since $4=4$,the point $\left(3, \frac{1}{2}\right)$ lies on the parabola.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

$A$ point on the parabola whose focus is $S(1,-1)$ and whose vertex is $A(1,1)$ is

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