Question:

The eccentricity of the ellipse $9x^2 + 25y^2 = 225$ is

Updated On: May 14, 2024
  • $\frac{3}{4}$
  • $\frac{4}{5}$
  • $\frac{9}{16}$
  • $\frac{3}{5}$
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The Correct Option is B

Solution and Explanation

$9x^{2}+25y^{2}=225$
$i.e., \frac{9 x^{2}}{225}+\frac{25y^{2}}{225}=1$
$\Rightarrow\, \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
$\Rightarrow\, \frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$
$a=5, b=3$
$c=\sqrt{a^{2}-b^{2}}=\sqrt{5^{2}-3^{2}}=\sqrt{16}=4$
$Eccentricity \, e=\frac{c}{a}=\frac{4}{5}$
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Concepts Used:

Ellipse

Ellipse Shape

An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity

Properties 

  • Ellipse has two focal points, also called foci.
  • The fixed distance is called a directrix.
  • The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
  • The total sum of each distance from the locus of an ellipse to the two focal points is constant
  • Ellipse has one major axis and one minor axis and a center

Read More: Conic Section

Eccentricity of the Ellipse

The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.

The eccentricity of ellipse, e = c/a

Where c is the focal length and a is length of the semi-major axis.

Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]

Area of an ellipse

The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.

Position of point related to Ellipse

Let the point p(x1, y1) and ellipse

(x2 / a2) + (y2 / b2) = 1

If [(x12 / a2)+ (y12 / b2) − 1)]

= 0 {on the curve}

<0{inside the curve}

>0 {outside the curve}