Q116 / 180 · KCET 2007 · Mathematics
The point on the curve $y^{2}=x$, the tangent at which makes an angle $45^{\circ}$ with $X$-axis is
- Correct answerA$\left(\frac{1}{4}, \frac{1}{2}\right)$
- B$\left(\frac{1}{2}, \frac{1}{4}\right)$
- C$\left(\frac{1}{2}, \frac{-1}{2}\right)$
- D$\left(\frac{1}{2}, \frac{1}{2}\right)$
Solution
Given,
$$
y^{2}=x
$$
$$
\begin{aligned}
&\Rightarrow \quad 2 y \frac{d y}{d x}=1 \\
&\Rightarrow \quad \frac{d y}{d x}=\frac{1}{2 y}=\text { Slope }
\end{aligned}
$$
Also given, $\quad \theta=45^{\circ}$
$\therefore$ Slope $\tan 45^{\circ}=1 \Rightarrow \frac{1}{2 y}+1$
$\Rightarrow \quad y=\frac{1}{2}$
From Eq. (i), if $y=\frac{1}{2}$, then $x=\frac{1}{4}$
$\therefore$ Required point is $\left(\frac{1}{4}, \frac{1}{2}\right)$.
Hence, the correct answer is option: 'A'.