Focal Distance of a Point on the Ellipse

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.





11 and 12 Grade Math

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