PSQ is a focal chord of the ellipse `4x^2+9y^2=36` such that SP=4. If S' the another focus write the value of S'Q.

Options

A. `2/4`
B. `15/4`
C. `7`
D. `26/5`

Explanation

To solve the problem step by step, we will follow the reasoning used in the video transcript while providing clear explanations for each step.

### Step 1: Write the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ 4x^2 + 9y^2 = 36 \]

To convert it into standard form, we divide both sides by 36:
\[ \frac{4x^2}{36} + \frac{9y^2}{36} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]

### Step 2: Identify the values of \(a\) and \(b\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 9\) → \(a = 3\)
- \(b^2 = 4\) → \(b = 2\)

### Step 3: Find the foci of the ellipse
The foci of the ellipse are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 - b^2}\):
\[ c = \sqrt{9 - 4} = \sqrt{5} \]
Thus, the foci are at:
\[ S = (\sqrt{5}, 0) \quad \text{and} \quad S' = (-\sqrt{5}, 0) \]

### Step 4: Use the property of focal chords
We know that \(PSQ\) is a focal chord, and we are given that \(SP = 4\). The segments of the focal chord are in harmonic progression.

Let:
- \(PS = a\)
- \(SQ = c\)
- The semi-latus rectum \(b' = \frac{b^2}{a} = \frac{4}{3}\)

### Step 5: Set up the harmonic mean relationship
Using the harmonic mean property, we have:
\[ \frac{4}{3} = \frac{2ac}{a + c} \]
Substituting \(a = 4\):
\[ \frac{4}{3} = \frac{2 \cdot 4 \cdot c}{4 + c} \]

### Step 6: Solve for \(c\)
Cross-multiplying gives:
\[ 4 + c = \frac{8c}{\frac{4}{3}} \]
This simplifies to:
\[ 4 + c = 6c \]
Rearranging gives:
\[ 4 = 5c \]
Thus,
\[ c = \frac{4}{5} \]

### Step 7: Find \(S'Q\)
Using the property that \(SQ + S'Q = 2a\):
\[ SQ + S'Q = 2 \cdot 3 = 6 \]
Substituting \(SQ = \frac{4}{5}\):
\[ \frac{4}{5} + S'Q = 6 \]
Thus,
\[ S'Q = 6 - \frac{4}{5} = \frac{30}{5} - \frac{4}{5} = \frac{26}{5} \]

### Final Answer
The value of \(S'Q\) is:
\[ \boxed{\frac{26}{5}} \]

---