JEE Mains · Maths · STD 12 - 6. 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 \(\sqrt {\frac{{19}}{2}} \)
- C \(\sqrt {\frac{{15}}{2}} \)
- D \(\frac{{\sqrt {19} }}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{{\sqrt {15} }}{2}\)
Step-by-step Solution
Detailed explanation
\(D=\sqrt{\alpha^{2}+\left(\alpha^{2}-4\right)^{2}}\) \({D^2} = {\alpha ^2} + {\alpha ^4} + 16 - 8{\alpha ^2}\) \(=\alpha^{4}-7 \alpha^{2}+16\) \(\frac{d D^{2}}{d \alpha}=4 \alpha^{3}-14 \alpha=0\) \(2 \alpha\left(2 \alpha^{2}-7\right)=0\) \({\alpha ^2} = \frac{7}{2}\)…
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