Focal Distance of a Point on the Hyperbola

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

The sum of the focal distance of any point on a hyperbola is constant and equal to the length of the transverse axis of the hyperbola.

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

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 (CN - CK)

⇒ SP = e(x - \(\frac{a}{e}\))

⇒ SP = ex - a ………………..…….. (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, S'P - SP = (a + ex) - (ex - a) = a + ex - ex + a = 2a = transverse axis.

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

Note: This property leads to an alternative definition of hyperbola 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 a hyperbola and the two fixed points are the two foci of the hyperbola.

Solved example to find the focal distance of any point on a hyperbola:

Find the focal distance of a point on the hyperbola 9x\(^{2}\) - 16y\(^{2}\) - 18x + 32y - 151 = 0.

Solution:

The given equation of the hyperbola is 9x\(^{2}\) - 16y\(^{2}\) - 18x + 32y - 151 = 0.

From the above equation we get,

9x\(^{2}\) - 18x - 16y\(^{2}\) + 32y = 151

⇒ 9(x\(^{2}\) - 2x) - 16(y\(^{2}\) - 2y) = 151

⇒ 9(x\(^{2}\) - 2x + 1) - 16(y\(^{2}\) - 2y + 1) = 151 + 9 - 16

⇒ 9(x - 1)\(^{2}\) - 16(y - 1)\(^{2}\) = 144

⇒ \(\frac{(x - 1)^{2}}{16}\) - \(\frac{(y - 1)^{2}}{9}\) = 1 ………………….. (i)

Now transfering the origin at (1, 1) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by x and y, we have

x = X + 1 and y = Y + 1 ………………….. (ii)

Using these relations, equation (i) reduces to

\(\frac{X^{2}}{4^{2}}\) - \(\frac{Y^{2}}{3^{2}}\) = 1 ……………………… (iii)

This is the form of \(\frac{X^{2}}{a^{2}}\) - \(\frac{Y^{2}}{b^{2}}\) = 1 (a\(^{2}\) > b\(^{2}\) ) where a = 4 and b = 3

Now, we get that a > b.

Hence, the equation\(\frac{X^{2}}{4^{2}}\) - \(\frac{Y^{2}}{3^{2}}\) = 1 represents an hyperbola whose transverse axes along X and conjugate axes along Y axes.

Therefore, the focal distance of a point on the hyperbola 9x\(^{2}\) - 16y\(^{2}\) - 18x + 32y - 151 = 0 is transverse axis = 2a = 2 ∙ 4 = 8 units.

11 and 12 Grade Math

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