KCET · Maths · Vector Algebra
The point on the curve \( y^{2}=x \) where the tangent makes an angle of \( \pi / 4 \) with \( x \)-axis is
- A \( \left(\frac{1}{2}, \frac{1}{4}\right) \)
- B \( \left(\frac{1}{4}, \frac{1}{2}\right) \)
- C \( (4,2) \)
- D \( (1,1) \)
Answer & Solution
Correct Answer
(B) \( \left(\frac{1}{4}, \frac{1}{2}\right) \)
Step-by-step Solution
Detailed explanation
Given that, \( y^{2}=x \rightarrow(1) \)
Slope of the curve is given by
\( m=\frac{d y}{d x} \)
And we know that, \( m=\tan \theta \)
Here, \( \theta=\frac{I I}{4} \) so, \( m=\tan \left(\frac{I}{4}\right)=1 \)
Now, differentiating Eq. (1) with respect to \( x \), we get
\( 2 y \frac{d y}{d x}=1 \) \( \Rightarrow 2 y \cdot 1=1 \) \( \Rightarrow y=\frac{1}{2} \) So, \( x=\frac{1}{4} \)
Therefore, point is \( \left(\frac{1}{4}, \frac{1}{2}\right) \)
Slope of the curve is given by
\( m=\frac{d y}{d x} \)
And we know that, \( m=\tan \theta \)
Here, \( \theta=\frac{I I}{4} \) so, \( m=\tan \left(\frac{I}{4}\right)=1 \)
Now, differentiating Eq. (1) with respect to \( x \), we get
\( 2 y \frac{d y}{d x}=1 \) \( \Rightarrow 2 y \cdot 1=1 \) \( \Rightarrow y=\frac{1}{2} \) So, \( x=\frac{1}{4} \)
Therefore, point is \( \left(\frac{1}{4}, \frac{1}{2}\right) \)
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