The axis of a parabola is the line y=x and it | vedclass

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(B) The vertex $V$ is at a distance $\sqrt{2}$ from the origin on the line $y=x$,so $V = (1, 1)$.The focus $S$ is at a distance $2\sqrt{2}$ from the o

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$9$

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(B) The vertex $V$ is at a distance $\sqrt{2}$ from the origin on the line $y=x$,so $V = (1, 1)$.The focus $S$ is at a distance $2\sqrt{2}$ from the origin on the line $y=x$,so $S = (2, 2)$.The distance $a$ between the vertex and the focus is $a = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$.The directrix is perpendicular to the axis $y=x$ and passes through a point $Z$ such that $V$ is the midpoint of $SZ$. Since $S=(2,2)$ and $V=(1,1)$,$Z$ must be $(0,0)$.The equation of the directrix is $x+y=0$.By the definition of a parabola,the distance from any point $P(1, k)$ on the parabola to the focus $S(2, 2)$ equals the distance from $P$ to the directrix $x+y=0$.$PS = \sqrt{(1-2)^2 + (k-2)^2} = \sqrt{1 + (k-2)^2}$.$PM = \frac{|1+k|}{\sqrt{1^2+1^2}} = \frac{|1+k|}{\sqrt{2}}$.Equating $PS^2 = PM^2$:$1 + (k-2)^2 = \frac{(1+k)^2}{2}$$2(1 + k^2 - 4k + 4) = 1 + k^2 + 2k$$2k^2 - 8k + 10 = 1 + k^2 + 2k$$k^2 - 10k + 9 = 0$$(k-1)(k-9) = 0$.Thus,$k=1$ or $k=9$. Since $k=1$ corresponds to the vertex,the other possible value is $k=9$.

The axis of a parabola is the line $y=x$ and its vertex and focus are in the first quadrant at distances $\sqrt{2}$ and $2\sqrt{2}$ units from the origin,respectively. If the point $(1, k)$ lies on the parabola,then a possible value of $k$ is :-

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