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Standard XII

Mathematics

General Equation of Ellipse

The parabolas...

Question

The parabolas $y^{2} = 4 x, x^{2} = 4 y$ divide the square region bounded by the lines $x = 4, y = 4$ and the coordinates axes. If $A_{1}, A_{2}, A_{3}$ are respectively the areas of these parts numbered top to bottom then $\frac{A_{1}}{A_{2}} + \frac{A_{2}}{A_{3}} + \frac{A_{3}}{A_{1}} =$ .......

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Solution

The correct option is D $3$
Area A_2 is common area between parabolas

$\Rightarrow A_{2} = \int_{0}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) d x$

$\Rightarrow 2 [\frac{x^{3 / 2}}{3 / 2}]_{0}^{4} - [\frac{x^{3}}{12}]_{0}^{4}$

$\Rightarrow 2 \times \frac{4^{3 / 2}}{3 / 2} - \frac{4^{3}}{12}$

$\Rightarrow \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$

Therefore, $A_{2} = \frac{16}{3}$

Now, $A_{1} + A_{3} = (4 \times 4) - \frac{16}{3}$

$\Rightarrow A_{1} + A_{3} = 16 - \frac{16}{3} = \frac{32}{3}$

$\Rightarrow A_{1} = A_{3} = \frac{1}{2} \times \frac{32}{3} = \frac{16}{3}$

Therefore, $A_{1} = A_{2} = A_{3} = \frac{16}{3}$

Hence, $\frac{A_{1}}{A_{2}} + \frac{A_{2}}{A_{3}} + \frac{A_{3}}{A_{1}} = 1 + 1 + 1 = 3$

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