Question

A point P on an ellipse is at a distance 6 units from a focus. If the eccentricity of the ellipse is \( \frac{3}{5} \), then the distance of P from the corresponding directrix is:

• A. \( \frac{8}{5} \)
• B. \( \frac{5}{8} \)
• C. \( 10 \)
• D. \( 12 \)
• E. \( 15 \)

Hint:

The focus-directrix property applies to all conics. Just remember the mnemonic: "Focus distance is $e$ times Directrix distance" ($PS = e \cdot PM$).

Answer 1

The Correct Option is C

Concept: The fundamental definition of a conic section (ellipse, parabola, or hyperbola) is based on the constant ratio called eccentricity (\(e\)). For any point \(P\) on the conic, the ratio of its distance from the focus (\(S\)) to its perpendicular distance from the corresponding directrix (\(M\)) is constant and equal to \(e\). \[ \frac{PS}{PM} = e \]

Step 1: Identify the given values.
Distance from the focus, \( PS = 6 \). Eccentricity of the ellipse, \( e = \frac{3}{5} \). We need to find the distance from the directrix, \( PM \).

Step 2: Apply the eccentricity formula.
From the definition \( PS = e \cdot PM \), we can rearrange to solve for \( PM \): \[ PM = \frac{PS}{e} \]

Step 3: Substitute the values and calculate.
\[ PM = \frac{6}{3/5} \] \[ PM = 6 \times \frac{5}{3} \] \[ PM = 2 \times 5 = 10 \]