JEE Mains · Maths · STD 12 - 8. Application and integration
If the area enclosed between the curves \(y = kx^2\) and \(x = ky^2, (k > 0)\), is \(1\) square unit. Then \(k\) is
- A \(\frac{{\sqrt 3 }}{2}\)
- B \(\frac{1}{{\sqrt 3 }}\)
- C \(\sqrt 3 \)
- D \(\frac{2}{{\sqrt 3 }}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{{\sqrt 3 }}\)
Step-by-step Solution
Detailed explanation
\(y=k x^{2}, x=k y^{2}\) \(\Rightarrow x=k\left(k^{2} x^{4}\right) \) \(\Rightarrow x=0\) or \(x^{3}=\left(\frac{1}{k}\right)^{3}\) \( \Rightarrow x=\frac{1}{k}, 0\) Point of intersection are \(\left(\frac{1}{\mathrm{k}}, \frac{1}{\mathrm{k}}\right)\) and \((0,0)\) Area…
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