Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #48

Using integration, find the area of the triangle formed by the negative x-axis and the tangent and normal to x²+y²=9 at (−1,2√2).

✅ Step-by-Step Solution

Final Answer: \(9 \sqrt{2}\) square units

Implicit differentiation gives y′=−x/y. At P=(−1,2√2), the tangent slope is 1/(2√2) and normal slope is −2√2. The tangent is y=(x+9)/(2√2), meeting the negative x-axis at (−9,0); the normal is y=−2√2x, meeting it at the origin. The top boundary changes at x=−1.
Use upper minus lower strip widths: \(\frac{\sqrt{2} x}{4} + \frac{9 \sqrt{2}}{4}\) for x∈\(\left[-9, -1\right]\); \(- 2 \sqrt{2} x\) for x∈\(\left[-1, 0\right]\).
Hence A=\(\int_{-9}^{-1}(\frac{\sqrt{2} x}{4} + \frac{9 \sqrt{2}}{4})\,dx\) + \(\int_{-1}^{0}(- 2 \sqrt{2} x)\,dx\).
Primitive \(\frac{\sqrt{2} x^{2}}{8} + \frac{9 \sqrt{2} x}{4}\), endpoint difference \(8 \sqrt{2}\); Primitive \(- \sqrt{2} x^{2}\), endpoint difference \(\sqrt{2}\).
Therefore the area is \(9 \sqrt{2}\) square units.

[1 mark] Implicit differentiation gives y′=−x/y. At P=(−1,2√2), the tangent slope is 1/(2√2) and normal slope is −2√2. The tangent is y=(x+9)/(2√2), meeting the negative x-axis at (−9,0); the normal is y=−2√2x, meeting it at the origin. The top boundary changes at x=−1.
[1 mark] Use upper minus lower strip widths: \(\frac{\sqrt{2} x}{4} + \frac{9 \sqrt{2}}{4}\) for x∈\(\left[-9, -1\right]\); \(- 2 \sqrt{2} x\) for x∈\(\left[-1, 0\right]\).
[1 mark] Hence A=\(\int_{-9}^{-1}(\frac{\sqrt{2} x}{4} + \frac{9 \sqrt{2}}{4})\,dx\) + \(\int_{-1}^{0}(- 2 \sqrt{2} x)\,dx\).
[1 mark] Primitive \(\frac{\sqrt{2} x^{2}}{8} + \frac{9 \sqrt{2} x}{4}\), endpoint difference \(8 \sqrt{2}\); Primitive \(- \sqrt{2} x^{2}\), endpoint difference \(\sqrt{2}\).
[1 mark] Therefore the area is \(9 \sqrt{2}\) square units.

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