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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