AP EAMCET · Maths · Application of Derivatives
The minimum distance of a point on the curve \(y=x^2-4\) from the origin is
- A \(\frac{\sqrt{15}}{2}\)
- B \(\frac{\sqrt{19}}{2}\)
- C \(\sqrt{\frac{15}{2}}\)
- D \(\sqrt{\frac{19}{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{15}}{2}\)
Step-by-step Solution
Detailed explanation
Curve is \(y=x^2-4\). Minimum distance from origin is along the normal passing through origin. Consider a point on curve \(\left(h, h^2-4\right)\). Slope of tangent is \(2 x=2 h\) at \(\left(h, h^2-4\right)\) \(\therefore\) Slope of normal is \(-\frac{1}{2 h}\). \(\therefore\)…
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