MHT CET · Maths · Application of Derivatives
The shortest distance between the line \(y-x=1\) and the curve \(x=y^2\) is
- A \(\frac{3 \sqrt{2}}{8}\)
- B \(\frac{2 \sqrt{3}}{8}\)
- C \(\frac{3 \sqrt{2}}{5}\)
- D \(\frac{\sqrt{3}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \sqrt{2}}{8}\)
Step-by-step Solution
Detailed explanation
Slope of line \(x-y+1=0\) is \(m_L = 1\). For curve \(x=y^2\), \(\frac{dx}{dy}=2y\). Slope of tangent \(\frac{dy}{dx} = \frac{1}{2y}\).
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