Question

If the length of the transverse and conjugate axes of a hyperbola be 8 and 6 respectively, then the difference of focal distances of any point of the hyperbola will be​

Answer 1

Step-by-step explanation:

2a = 8,2b = 6

difference of focal distances of any point of hyperbola

2a = 8

Answer 2

The difference of focal distances of the point (4√2,3) on the hyperbola is 8 units.

Step-by-step explanation:

If the equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, then it's conjugate axis length is 2a = 8 (Given), ⇒ a = 4 and its transverse axis length is 2b = 6 (Given), ⇒ b = 3

Now, the equation of the hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$.

The eccentricity of the hyperbola, e = $\sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{3^{2} }{4^{2}}} = \frac{5}{4}$

Now, the coordinates of the foci are (± ae, 0) = (± 5, 0)

Now, (4√2, 3) is a point on the hyperbola.

So, the difference of focal distances of the point (4√2,3) on the hyperbola is $[\sqrt{(4\sqrt{2} + 5)^{2} + (3)^{2} }] - [\sqrt{(4\sqrt{2} - 5)^{2} + (3)^{2} }]$

= 11.07 - 3.07

= 8 units. (Answer)