Question

Let \( P \) be a point on an ellipse at a distance of \( 8 \) units from a focus. If the eccentricity is \( \frac{4}{5} \), then the distance of the point \( P \) from the directrix is:

• A. $\frac{5}{8}$
• B. $\frac{8}{5}$
• C. $5$
• D. $8$
• E. $10$

Hint:

Remember the Focus-Directrix property: $PS = e \cdot PM$. For an ellipse, $e < 1$, so the distance to the directrix will always be greater than the distance to the focus.

Answer 1

Answer: (E)

Concept: The fundamental definition of a conic section (including an ellipse) is the ratio of the distance of a point from the focus ($PS$) to its distance from the directrix ($PM$). This ratio is the eccentricity ($e$): \[ \frac{PS}{PM} = e \]

Step 1: Identify the given values.

• Distance from focus ($PS$) = $8$ units.
• Eccentricity ($e$) = $4/5$.

Step 2: Set up the ratio formula.
Let $d$ be the distance from the point $P$ to the directrix. \[ \frac{8}{d} = \frac{4}{5} \]

Step 3: Solve for $d$.
Cross-multiply to find $d$: \[ 4d = 8 \times 5 \] \[ 4d = 40 \Rightarrow d = 10 \]