KCET · Maths · Application of Derivatives
The point on the curve \(y^{2}=x\), the tangent at which makes an angle \(45^{\circ}\) with \(X\)-axis is
- A \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
- B \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
- C \(\left(\frac{1}{2}, \frac{-1}{2}\right)\)
- D \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
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 \(\left(\frac{1}{4}, \frac{1}{2}\right)\).
\[
y^{2}=x
\]
\[
\begin{aligned}
&\Rightarrow \quad 2 y \frac{d y}{d x}=1 \\
&\Rightarrow \quad \frac{d y}{d x}=\frac{1}{2 y}=\text { Slope }
\end{aligned}
\]
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 \(\left(\frac{1}{4}, \frac{1}{2}\right)\).
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