Question

The eccentricity of the hyperbola which is the locus of a point moving in a plane such that the difference of its distances from the points \[ (-5,0) \quad\text{and}\quad (5,0) \] is equal to \(8\), is

• A. \(\dfrac54\)
• B. \(\dfrac45\)
• C. \(-\dfrac54\)
• D. \(-\dfrac45\)

Hint:

For a hyperbola: \[ \big|PF_1-PF_2\big|=2a \] and \[ e=\frac{c}{a}\gt 1. \] Eccentricity of a hyperbola is always positive and greater than \(1\).

Answer 1

The Correct Option is A

Step 1: Recognise the definition.
A point whose distances from two fixed points (the foci) have a constant difference traces a hyperbola. Here the foci are $(-5,0)$ and $(5,0)$ and the constant difference is $8$.

Step 2: Find $c$.
The foci sit at $(\pm c,0)$, so comparing with $(\pm5,0)$ gives $c=5$.

Step 3: Find $a$.
For a hyperbola the constant difference of focal distances equals $2a$. So $2a=8$, giving $a=4$.

Step 4: Recall the eccentricity formula.
For a hyperbola the eccentricity is $e=\dfrac{c}{a}$.

Step 5: Substitute.
$e=\dfrac{5}{4}$.

Step 6: Note the sign.
Eccentricity is always positive, so the answer is $\dfrac{5}{4}$, not the negative options. \[ \boxed{\dfrac{5}{4}} \]