Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(16x^2 + y^2 = 16\).

Updated On: Oct 25, 2023
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Solution and Explanation

The given equation is:
\(16x^2 + y^2 = 16\)
or \(\frac {x^2}{1}+ \frac {y^2}{16} = 1\)
or \(\frac {x^2}{12} + \frac {y^2}{42} = 1\) .......(1)
Here, the denominator of \(\frac {y^2}{4^2}\) is greater than the denominator of \(\frac {x^2}{1^2}\).
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation (1) with \(\frac {x^2}{b^2} + \frac {y^2}{a^2} = 1\), we obtain b = 1 and a = 4.
∴ \(c = \sqrt {(a^2 – b^2)}\)
\(c= \sqrt {(16-1)}\)
\(c= \sqrt {15}\)
Therefore, 
The coordinates of the foci are \((0, ±\sqrt {15})\)
The coordinates of the vertices are (0, ±4). 
Length of major axis = 2a = 8
Length of the minor axis = 2b = 2
Eccentricity, \(e = \frac ca = \frac {\sqrt {15}}{4}\)
Length of latus rectum = \(\frac {2b^2}{a} = \frac {(2×1^2)}{4} = \frac 24 = \frac 12\)

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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}