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
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
Show 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 \(1\).
Updated On: Jun 16, 2026
- \(\dfrac54\)
- \(\dfrac45\)
- \(-\dfrac54\)
- \(-\dfrac45\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For a hyperbola,
\[
\text{Difference of distances from the foci}
=
2a
\]
and
\[
e=\frac{c}{a}
\]
where \(c\) is the focal distance.
Step 1: Identify \(c\). The foci are \[ (-5,0) \quad\text{and}\quad (5,0) \] Hence \[ c=5. \]
Step 2: Use the given difference of distances. Given \[ 2a=8 \] Therefore, \[ a=4. \]
Step 3: Calculate the eccentricity. \[ e=\frac{c}{a} = \frac{5}{4} \] \[\begin{aligned} \boxed{\frac54} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
Step 1: Identify \(c\). The foci are \[ (-5,0) \quad\text{and}\quad (5,0) \] Hence \[ c=5. \]
Step 2: Use the given difference of distances. Given \[ 2a=8 \] Therefore, \[ a=4. \]
Step 3: Calculate the eccentricity. \[ e=\frac{c}{a} = \frac{5}{4} \] \[\begin{aligned} \boxed{\frac54} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
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