Q106 / 180 · COMEDK 2020 · Mathematics
The point on the curve $y^{2}=x$, the tangent at which makes an angle $45^{\circ}$ with $X$-axis is
- A$(1 / 2,1 / 4)$
- Correct answerB$(1 / 4,1 / 2)$
- C$(1 / 2,1 / 2)$
- D$(1 / 2,-1 / 2)$
Solution
We have,
$$
\begin{gathered}
y^{2}=x \\
\Rightarrow 2 y \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}
\end{gathered}
$$
Let the required point be $\left(x_{1}, y_{1}\right)$, So, slope of tangent
$$
\begin{aligned}
&=\left.\frac{d y}{d x}\right|_{\left(x_{1}-w_{1}\right)} \\
&\Rightarrow \quad m=\frac{1}{2 y_{1}}
\end{aligned}
$$
Since, tangent makes $45^{5}$ with $X$-axis, so
$$
m=\tan 45^{\circ}=1
$$
From Eqs, (i) and (ii), we have
$$
\frac{1}{2 q_{1}}=1 \Rightarrow f_{1}=\frac{1}{2}
$$
Again, $\left(x_{1}, y_{1}\right)$ lies on th curve $y^{Z}=x_{1}$ So,
$$
y_{1}^{2}=x_{1} \Rightarrow x_{1}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}
$$
So, required point is $\left(\frac{1}{4}, \frac{1}{2}\right)$.
Hence, the correct answer is option: 'B'.