Q:Ellipse x²/16 + y²/9 = 1 Area = ?

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For the ellipse x²/16 + y²/9 = 1 Here, a = 4 and b = 3. Area = πab = π × 4 × 3 = 12π square units ≈ 37.7 square units.
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Center: $(0, 0)$ Orientation: Horizontal major axis (along the x-axis, as $16 > 9$) Semi-Major Axis ($a$): $4$ (since $a^2 = 16$) Semi-Minor Axis ($b$): $3$ (since $b^2 = 9$) Focal Distance ($c$): $\sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7} \approx 2.65$ Major Axis...
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Center: $(0, 0)$ Orientation: Horizontal major axis (along the x-axis, as $16 > 9$) Semi-Major Axis ($a$): $4$ (since $a^2 = 16$) Semi-Minor Axis ($b$): $3$ (since $b^2 = 9$) Focal Distance ($c$): $\sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7} \approx 2.65$ Major Axis Vertices: $(\pm 4, 0)$ Minor Axis Vertices (Co-vertices): $(0, \pm 3)$ Foci: $(\pm\sqrt{7}, 0)$ Length of Major Axis: $2a = 8$ Length of Minor Axis: $2b = 6$ Eccentricity ($e$): $\frac{c}{a} = \frac{\sqrt{7}}{4} \approx 0.661$ Directrices: $x = \pm\frac{a^2}{c} = \pm\frac{16}{\sqrt{7}}$ Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2} = 4.5$ read less
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Find the area of the region bounded by the ellipse (x2/16) + (y2/9) = 1. A. The given ellipse (x2/16) + (y2/9)=1 can be represented as given below: Ellipse is symmetrical about x-axis and y-axis.So, area bounded by ellipse = 4 x Area of OAB Areaof OAB=∫04ydx     =∫0431−x216dx     =34∫0416−x2dx     =3404     =34−34     =34×8π2     =3π square unitsTherefore,area...
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Find the area of the region bounded by the ellipse (x2/16) + (y2/9) = 1. A. The given ellipse (x2/16) + (y2/9)=1 can be represented as given below: Ellipse is symmetrical about x-axis and y-axis.So, area bounded by ellipse = 4 x Area of OAB Areaof OAB=∫04ydx     =∫0431−x216dx     =34∫0416−x2dx     =34[x216−x2+162sin−1x4]04     =34[4216−42+162sin−144]−34[0216−02+162sin−104]     =34×8π2     =3π square unitsTherefore,area bounded by ellipse     =4×Areaof OAB     =4×3π square units     =12 π square units read less
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Professional experience with 7 years of experience in Corporate industry

12π sq units
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Here is the complete geometric analysis of the ellipse given by the equation: $$\frac{x^2}{16} + \frac{y^2}{9} = 1$$ 📊 Key Geometric Properties Standard Form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Orientation: Horizontal major axis (since $a^2 > b^2$) Semi-major axis ($a$): $4$ (since...
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Here is the complete geometric analysis of the ellipse given by the equation: $$\frac{x^2}{16} + \frac{y^2}{9} = 1$$ 📊 Key Geometric Properties Standard Form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Orientation: Horizontal major axis (since $a^2 > b^2$) Semi-major axis ($a$): $4$ (since $a^2 = 16$) Semi-minor axis ($b$): $3$ (since $b^2 = 9$) 🔍 Detailed Components Center: $(0, 0)$ Vertices (on the major axis): $(\pm 4, 0)$ Co-vertices (on the minor axis): $(0, \pm 3)$ Foci: First, find the linear eccentricity ($c$) using $c^2 = a^2 - b^2$:$$c^2 = 16 - 9 = 7 \implies c = \sqrt{7} \approx 2.65$$ Foci coordinates: $(\pm \sqrt{7}, 0)$ Eccentricity ($e$):$$e = \frac{c}{a} = \frac{\sqrt{7}}{4} \approx 0.66$$ Directrices:$$x = \pm \frac{a^2}{c} = \pm \frac{16}{\sqrt{7}}$$ Length of Major Axis: $2a = \mathbf{8}$ Length of Minor Axis: $2b = \mathbf{6}$ Length of Latus Rectum:$$\frac{2b^2}{a} = \frac{2(9)}{4} = \mathbf{4.5}$$ read less
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Professional experience with 7 years of experience in Corporate industry

12π ≈ 37.70 sq. units
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Maths wizard

π*4*3=12π
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