Question:

Step 1: Identify key properties from the given information. Let $S(0,0)$ be the nearer focus and $P(3,4)$ be a point on the ellipse. The eccentricity is $e = \frac{1}{2}$. The tangent to the ellipse at $P$ is given by the equation $3x + 4y - 25 = 0$. First, calculate the distance from the focus $S$ to the point $P$: $SP = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Next, determine the relationship between the line segment $SP$ and the tangent at $P$. The slope of the line $SP$ is $m_{SP} = \frac{4-0}{3-0} = \frac{4}{3}$. The slope of the tangent $3x+4y-25=0$ is $m_{tangent} = -\frac{3}{4}$. Since $m_{SP} \cdot m_{tangent} = \frac{4}{3} \cdot \left(-\frac{3}{4}\right) = -1$, the line segment $SP$ is perpendicular to the tangent at $P$. This implies that $P$ is the foot of the perpendicular from the focus $S$ to the tangent. Step 2: Apply the reflection property of the ellipse. Let $S_1$ be the reflection of the focus $S(0,0)$ across the tangent $3x+4y-25=0$. Since $P(3,4)$ is the foot of the perpendicular from $S$ to the tangent, $P$ is the midpoint of the segment $SS_1$. Therefore, $S_1 = 2P - S = 2(3,4) - (0,0) = (6,8)$. A fundamental property of an ellipse is that the reflection of one focus across any tangent lies on the circle centered at the other focus $S'$ with radius $2a$, where $a$ is the semi-major axis. Thus, the distance $S'S_1 = 2a$. Step 3: Set up equations for the semi-major axis $a$ and the other focus $S'$. Let the other focus be $S'=(x',y')$. From the reflection property, $S'S_1 = 2a$: $$ (x'-6)^2 + (y'-8)^2 = (2a)^2 \quad (1) $$ For any point $P$ on the ellipse, the sum of the distances from $P$ to the two foci is $2a$: $SP + S'P = 2a$. We found $SP=5$, so $S'P = 2a-5$. $$ (x'-3)^2 + (y'-4)^2 = (2a-5)^2 \quad (2) $$ The distance between the foci is $2c$. Since $S=(0,0)$ and $S'=(x',y')$, we have $x'^2+y'^2 = (2c)^2$. The eccentricity is $e = \frac{c}{a} = \frac{1}{2}$, which implies $c = \frac{a}{2}$, or $2c = a$. Substituting $2c=a$ into the distance between foci equation: $$ x'^2+y'^2 = a^2 \quad (3) $$ Step 4: Solve for $a$. Expand equation (2): $x'^2 - 6x' + 9 + y'^2 - 8y' + 16 = (2a-5)^2$ Substitute $x'^2+y'^2=a^2$ from equation (3): $a^2 - 6x' - 8y' + 25 = (2a-5)^2 \quad (4)$ Expand equation (1): $x'^2 - 12x' + 36 + y'^2 - 16y' + 64 = (2a)^2$ Substitute $x'^2+y'^2=a^2$ from equation (3): $a^2 - 12x' - 16y' + 100 = (2a)^2 \quad (5)$ From equation (4), rearrange to find $6x'+8y'$: $6x'+8y' = a^2 + 25 - (2a-5)^2 = a^2 + 25 - (4a^2 - 20a + 25) = -3a^2 + 20a$. Dividing by 2: $3x'+4y' = \frac{1}{2}(-3a^2+20a) \quad (6)$ From equation (5), rearrange to find $12x'+16y'$: $12x'+16y' = a^2 + 100 - (2a)^2 = a^2 + 100 - 4a^2 = -3a^2 + 100$. Dividing by 4: $3x'+4y' = \frac{1}{4}(-3a^2+100) \quad (7)$ Equating the expressions for $3x'+4y'$ from (6) and (7): $\frac{1}{2}(-3a^2+20a) = \frac{1}{4}(-3a^2+100)$ Multiply by 4: $2(-3a^2+20a) = -3a^2+100$ $-6a^2+40a = -3a^2+100$ $3a^2 - 40a + 100 = 0$. Solve the quadratic equation for $a$: $a = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(3)(100)}}{2(3)}$ $a = \frac{40 \pm \sqrt{1600 - 1200}}{6}$ $a = \frac{40 \pm \sqrt{400}}{6}$ $a = \frac{40 \pm 20}{6}$. This yields two possible values for $a$: $a_1 = \frac{40+20}{6} = \frac{60}{6} = 10$. $a_2 = \frac{40-20}{6} = \frac{20}{6} = \frac{10}{3}$. Step 5: Determine the correct value of $a$ and the ellipse parameters. The problem states that $S(0,0)$ is the *nearer* focus to $P(3,4)$. For $a=10$: $SP=5$. $S'P = 2a-SP = 2(10)-5 = 15$. Since $SP < S'P$ ($5 < 15$), $S$ is the nearer focus. This is consistent. For $a=10/3$: $SP=5$. $S'P = 2a-SP = 2(10/3)-5 = 20/3-15/3 = 5/3$. Since $SP > S'P$ ($5 > 5/3$), $S'$ would be the nearer focus. This contradicts the given information. Therefore, the correct semi-major axis is $a=10$. Now, calculate the other parameters: Semi-minor axis $c = ae = 10 \cdot \frac{1}{2} = 5$. $b^2 = a^2 - c^2 = 10^2 - 5^2 = 100 - 25 = 75$. Step 6: Calculate the focal length, length of latus rectum, and center of the ellipse. 1. **Focal length:** The focal length is $2c$. $$ 2c = 2(5) = 10 $$ 2. **Length of latus rectum:** The length of the latus rectum is $\frac{2b^2}{a}$. $$ \frac{2b^2}{a} = \frac{2(75)}{10} = \frac{150}{10} = 15 $$ 3. **Center of the ellipse:** To find the center, first find the coordinates of $S'$. Using $a=10$ in equation (6): $3x'+4y' = \frac{1}{2}(-3(10)^2+20(10)) = \frac{1}{2}(-300+200) = -50$. Using $a=10$ in equation (3): $x'^2+y'^2 = 10^2 = 100$. From $3x'+4y'=-50$, we have $y' = \frac{-50-3x'}{4}$. Substitute this into $x'^2+y'^2=100$: $x'^2 + \left(\frac{-50-3x'}{4}\right)^2 = 100$ $16x'^2 + (-50-3x')^2 = 1600$ $16x'^2 + (2500 + 300x' + 9x'^2) = 1600$ $25x'^2 + 300x' + 900 = 0$ Divide by 25: $x'^2 + 12x' + 36 = 0$ $(x'+6)^2 = 0 \implies x' = -6$. Substitute $x'=-6$ into $y' = \frac{-50-3x'}{4}$: $y' = \frac{-50-3(-6)}{4} = \frac{-50+18}{4} = \frac{-32}{4} = -8$. So the other focus is $S'(-6,-8)$. The center $C$ is the midpoint of $S(0,0)$ and $S'(-6,-8)$: $$ C = \left(\frac{0+(-6)}{2}, \frac{0+(-8)}{2}\right) = (-3,-4) $$ Step 7: Calculate the length of the chord $AB$. The chord through $S$ is parallel to the tangent at $P$. The tangent at $P$ is $3x+4y-25=0$. The chord through $S(0,0)$ parallel to this tangent is $3x+4y=0$. The major axis of the ellipse passes through the foci $S(0,0)$ and $S'(-6,-8)$. The slope of the major axis is $\frac{-8-0}{-6-0} = \frac{4}{3}$. The equation of the major axis is $y = \frac{4}{3}x$, or $4x-3y=0$. The slope of the chord $3x+4y=0$ is $-\frac{3}{4}$. Since the product of the slopes of the major axis and the chord is $\frac{4}{3} \cdot \left(-\frac{3}{4}\right) = -1$, the chord is perpendicular to the major axis. As the chord passes through the focus $S(0,0)$ and is perpendicular to the major axis, it is a latus rectum. The length of the latus rectum was calculated in Step 6 as 15. Therefore, the length of $AB$ is 15.