Area of the ellipse x²/a² + y²/b² = 1 = πabदीर्घवृत्त x²/a² + y²/b² = 1 का क्षेत्रफल = πabEllipse x²/a² + y²/b² = 1 ka area = πab

NCERT Example 2 (Fig 8.7, 8.8) · Class 12 Maths · Chapter 8 · interactive diagram

Parts and what they mean

• x-axis (x-अक्ष) — The horizontal number line; inputs x are read here.
• y-axis (y-अक्ष) — The vertical number line; outputs y = f(x) are read here.
• Origin O(0, 0) (मूल बिंदु O(0, 0)) — Where the two axes cross.
• Ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (दीर्घवृत्त \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1) — Symmetric about both axes, so area = 4 \times area of AOBA = \pi ab.
• Region AOBA (first quadrant) (क्षेत्र AOBA (प्रथम चतुर्थांश)) — Area = \frac{b}{a}\int_0^a \sqrt{a^2 - x^2}\,dx = \frac{\pi ab}{4}.
• Vertical strip (ऊर्ध्वाधर पट्टी) — From the ellipse, y = \frac{b}{a}\sqrt{a^2 - x^2} (positive root, since the strip is in the first quadrant).
• Vertex A(a, 0) (शीर्ष A(a, 0)) — End of the major axis on the right: upper limit x = a.
• Point B(0, b) (बिंदु B(0, b)) — Top end of the minor axis: upper limit y = b for horizontal strips.
• Vertex A′(-a, 0) (शीर्ष A′(-a, 0)) — Left end of the major axis.
• Point B′(0, -b) (बिंदु B′(0, -b)) — Bottom end of the minor axis.

Original diagram drawn by Simple Learning Hub, based on NCERT Example 2 (Fig 8.7, 8.8). Report a mistake