Question

Find the area bounded by the region 1. A={(x,y):x^2÷16+y^2÷9 <=1 <=x÷4+y÷3}​

Answer 1

$\huge{\bf{Solution:-}}$

To find area {(x, y): y2≤ 8x, x2 + y2≤9}

y2 = 8x ...(i)

x2 + y2 = 9 ...(ii)

On solving the equation (i) and (ii),

Or, x2 + 8x = 9

Or, x2 + 8x – 9 = 0

Or, (x + 9)(x – 1) = 0

Or, x = – 9 or x = 1

And when x = 1 then y = ±2√2

Equation (i) represents a parabola with vertex (0,0) and axis as x – axis, equation (ii) represents a circle with centre (0,0) and radius 3 units, so it meets area at (±3, 0), (0,±3).

Point of intersection of parabola and circle is (1,2√2) and (1, – 2√2).

Answer 2

$\huge{\underline{\mathbb{\red{SOLUTION}}}}$

FIND AREA

{(x, y): y2≤ 8x, x2 + y2≤9}

y2 = 8x ...(i)

x2 + y2 = 9 ...(ii)

On solving the equation (i) and (ii),

Or, x2 + 8x = 9

Or, x2 + 8x – 9 = 0

Or, (x + 9)(x – 1) = 0

Or, x = – 9 or x = 1

And when x = 1 then y = ±2√2

Equation (i) represents a parabola with vertex (0,0) and axis as x – axis, equation (ii) represents a circle with centre (0,0) and radius 3 units, so it meets area at (±3, 0), (0,±3).

Point of intersection of parabola and circle is (1,2√2) and (1, – 2√2)

$\huge{\red{\ddot{\smile}}}$

$\huge{\underline{\underline{\mathfrak{Thank you}}}}$