CUET · MATHS · PYQ PAPER 2023
Slope of the tangent to the parabola \(y^2=x+2\) at a point in the 1st quadrant and lying on the line \(y=x\) is:
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{3}\)
- D 2
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(y=x\) \(y^2=x+2\) \(x^2=x+2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2 \implies y=2\) (1st quadrant) \(2y \frac{dy}{dx}=1\) \(\frac{dy}{dx}=\frac{1}{2y}\) \(m=\frac{1}{2(2)}\) \(m=\frac{1}{4}\)
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