JEE Mains · Maths · STD 12 - 8. Application and integration
The sum of squares of all possible values of \(k\), for which area of the region bounded by the parabolas \(2 \mathrm{y}^2=\mathrm{kx}\) and \(\mathrm{ky}^2=2(\mathrm{y}-\mathrm{x})\) is maximum, is equal to :
- A \(1\)
- B \(4\)
- C \(5\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\( k y^2=2(y-x) \) \( 2 y^2=k x \) \( \text { Point of intersection } \rightarrow \) \( \mathrm{ky}^2=2\left(\mathrm{y}-\frac{2 \mathrm{y}^2}{\mathrm{k}}\right) \) \( \mathrm{y}=0 \quad \mathrm{ky}=2\left(1-\frac{2 \mathrm{y}}{\mathrm{k}}\right) \)…
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