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

Question

Accepted Answer

"$\left(x+\frac{1}{2}\right)^{2}=\left(y+\frac{1}{4}\right)$

$y=\left(x^{2}+x\right)$

$\tan ^{-1} \sqrt{\mathrm{x}(\mathrm{x}+1)}+\sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}=\pi / 2$

$0 \leq \mathrm{x}^{2}+\mathrm{x}+1 \leq 1$

$x^{2}+x \leq 0$

Also $x^{2}+x \geq 0$

$\therefore \mathrm{x}^{2}+\mathrm{x}=0 \Rightarrow \mathrm{x}=0,-1$

$\mathrm{S}$ contains 2 element."

Let $\mathrm{y}=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y=-\frac{1}{2}$. Then

$$S=\left\{x \in \mathbb{R}: \tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}$$ :

Solution

About this question

Drill 25 more like these. Every day. Free.

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

Solution

About this question

More Conic Sections questions

Drill 25 more like these. Every day. Free.

Let $\mathrm{y}=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y=-\frac{1}{2}$. Then

$$S=\left\{x \in \mathbb{R}: \tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}$$ :