JEE Mains · Maths · STD 12 - 8. Application and integration
Consider a curve \(y=y(x)\) in the first quadrant as shown in the figure. Let the area \(A_{1}\) is twice the area \(A _{2}\). Then the normal to the curve perpendicular to the line \(2 x -12 y =15\) does NOT pass through the point.
- A \((6,21)\)
- B \((8,9)\)
- C \((10,4)\)
- D \((12,15)\)
Answer & Solution
Correct Answer
(C) \((10,4)\)
Step-by-step Solution
Detailed explanation
Given that \(A _{1}=2 A _{2}\) from the graph \(A _{1}+ A _{2}= xy -8\) \(\frac{3}{2} A _{1}= xy -8\) \(A _{1}=\frac{2}{3} xy -\frac{16}{3}\) \(\int_{4}^{ x } f ( x ) dx =\frac{2}{3} xy -\frac{16}{3}\) \(f ( x )=\frac{2}{3}\left( x \frac{ d y}{ dx }+ y \right)\)…
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