Question:

The distance of the point $'\theta'$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ from a focus is

Updated On: Oct 3, 2024
  • $a(e+cos\,\theta)$
  • $a(e-cos\,\theta)$
  • $a(1+e\,cos\,\theta)$
  • $a(1+2e\,cos\,\theta)$
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The Correct Option is C

Solution and Explanation

Focal distance of any point.
$ P\left(x, y\right)$ on the ellipse is $= SP= a+ex $
Here $x=a\,cos\, \theta $
$\therefore SP = a+ea\,cos\,\theta $
$= a\left(1+e\, cos\,\theta\right)$
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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}