Let y=f(x) represent a parabola with focus le | vedclass

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(A) The parabola has focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y = -\frac{1}{2}$.Using the definition of a parabola,the distance from $(x, y

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contains exactly two elements

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(A) The parabola has focus $\left(-\frac{1}{2}, 0ight)$ and directrix $y = -\frac{1}{2}$.Using the definition of a parabola,the distance from $(x, y)$ to the focus equals the distance to the directrix:$\sqrt{\left(x + \frac{1}{2}ight)^2 + y^2} = \left|y + \frac{1}{2}ight|$Squaring both sides: $\left(x + \frac{1}{2}ight)^2 + y^2 = y^2 + y + \frac{1}{4}$$\left(x + \frac{1}{2}ight)^2 = y + \frac{1}{4} \Rightarrow y = x^2 + x = f(x)$.We are given $	an^{-1}\left(\sqrt{f(x)}ight) + \sin^{-1}\left(\sqrt{f(x)+1}ight) = \frac{\pi}{2}$.Let $u = \sqrt{f(x)}$. Then $	an^{-1}(u) + \sin^{-1}(\sqrt{u^2+1}) = \frac{\pi}{2}$.This implies $\sin^{-1}(\sqrt{u^2+1}) = \frac{\pi}{2} - 	an^{-1}(u) = \cot^{-1}(u) = \sin^{-1}\left(\frac{1}{\sqrt{u^2+1}}ight)$.Thus,$\sqrt{u^2+1} = \frac{1}{\sqrt{u^2+1}}$ $\Rightarrow u^2+1 = 1$ $\Rightarrow u^2 = 0$ $\Rightarrow f(x) = 0$.Since $f(x) = x^2+x$,we have $x^2+x = 0$,which gives $x(x+1) = 0$.Thus,$x = 0$ or $x = -1$.The set $S = \{0, -1\}$ contains exactly two elements.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y =-\frac{1}{2}$. Then $S=\left\{x \in R : \tan ^{-1}\left(\sqrt{f(x)}+\sin ^{-1}(\sqrt{f(x)+1})\right)=\frac{\pi}{2}\right\}$:

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