What is the focal distance of a point on the ellipse?

The sum of the focal distance of any point on an ellipse is constant and equal to the length of the major axis of the ellipse.

Let P (x, y) be any point on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1.

Let MPM' be the perpendicular through P on directrices ZK and Z'K'. Now by definition we get,

SP = e **∙** PM

⇒ SP = e** ∙** NK

⇒ SP = e (CK - CN)

⇒ SP = e(\(\frac{a}{e}\) - x)

⇒ SP = a - ex ………………..…….. (i)

and

S'P = e **∙** PM'

⇒ S'P = e **∙** (NK')

⇒ S'P = e (CK' + CN)

⇒ S'P = e (\(\frac{a}{e}\) + x)

⇒ S'P = a + ex ………………..…….. (ii)

Therefore, SP + S'P = a - ex + a + ex = 2a = major axis.

Hence, the sum of the focal distance of a point P (x, y) on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 is constant and equal to the length of the major axis (i.e., 2a) of the ellipse.

**Note:** This
property leads to an
alternative definition of ellipse as follows:

If a point moves on a plane in such a way that the sum of its distances from two fixed points on the plane is always a constant then the locus traced out by the moving point on the plane is called an ellipse and the two fixed points are the two foci of the ellipse.

Solved example to find the focal distance of any point on an ellipse:

Find the focal distance of a point on the ellipse 25x\(^{2}\) + 9y\(^{2}\) -150x – 90y + 225 = 0

**Solution:**

The given equation of the ellipse is 25x\(^{2}\) + 9y\(^{2}\) - 150x - 90y + 225 = 0.

From the above equation we get,

25x\(^{2}\) - 150x + 9y\(^{2}\) - 90y = - 225

⇒ 25(x\(^{2}\) - 6x) + 9(y\(^{2}\) - 10y) = -225

⇒ 25(x\(^{2}\) - 6x + 9) + 9(y\(^{2}\) - 10y + 25) = 225

⇒ 25(x - 3)\(^{2}\) + 9(y - 5)\(^{2}\) = 225

⇒ \(\frac{(x - 3)^{2}}{9}\) + \(\frac{(y - 5)^{2}}{25}\) = 1 ………………….. (i)

Now transfering the origin at (3, 5) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by x and y, we have

x = X + 3 and y = Y + 5 ………………….. (ii)

Using these relations, equation (i) reduces to

\(\frac{X^{2}}{3^{2}}\) + \(\frac{Y^{2}}{5^{2}}\) = 1 ……………………… (iii)

This is the form of \(\frac{X^{2}}{b^{2}}\) + \(\frac{Y^{2}}{a^{2}}\) = 1 (a\(^{2}\) < b\(^{2}\) ) where a = 5 and b = 3

Now, we get that a > b.

Hence, the equation\(\frac{X^{2}}{3^{2}}\) + \(\frac{Y^{2}}{5^{2}}\) = 1 represents an ellipse
whose major axes along X and minor axes along Y axes.

Therefore, the focal distance of a point on the ellipse
25x\(^{2}\) + 9y\(^{2}\) - 150x - 90y + 225 = 0 is major axis = 2a = 2 **∙** 5 = 10 units.

**● ****The Ellipse**

**Definition of Ellipse****Standard Equation of an Ellipse****Two Foci and Two Directrices of the Ellipse****Vertex of the Ellipse****Centre of the Ellipse****Major and Minor Axes of the Ellipse****Latus Rectum of the Ellipse****Position of a Point with respect to the Ellipse****Ellipse Formulae****Focal Distance of a Point on the Ellipse****Problems on Ellipse**

**From Focal Distance of a Point on the Ellipse**** to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.