Question:
Let $R(-2,-2)$ be a point and let $\dfrac{(x-3)^2}{25} + \dfrac{(y+2)^2}{16} = 1$ be an ellipse. If $S$ and $T$ are the foci of the ellipse, then $RS + RT$ is equal to:
Let $R(-2,-2)$ be a point and let $\dfrac{(x-3)^2}{25} + \dfrac{(y+2)^2}{16} = 1$ be an ellipse. If $S$ and $T$ are the foci of the ellipse, then $RS + RT$ is equal to:
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For ellipse, sum of distances from any point on ellipse to foci is constant = $2a$.
Updated On: Apr 24, 2026
- $128$
- $61$
- $12$
- $10$
- $124$
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The Correct Option is D
Solution and Explanation
Concept:
• In ellipse, sum of distances from any point on ellipse to foci = $2a$
Step 1: Identify ellipse parameters
\[ \frac{(x-3)^2}{25} + \frac{(y+2)^2}{16} = 1 \] \[ a^2 = 25 \Rightarrow a = 5 \]
Step 2: Check if point lies on ellipse
For $R(-2,-2)$: \[ \frac{(-2-3)^2}{25} + \frac{(-2+2)^2}{16} = \frac{25}{25} + 0 = 1 \] So, $R$ lies on ellipse.
Step 3: Apply property
\[ RS + RT = 2a = 2 \times 5 = 10 \] Final Conclusion:
\[ = 10 \]
• In ellipse, sum of distances from any point on ellipse to foci = $2a$
Step 1: Identify ellipse parameters
\[ \frac{(x-3)^2}{25} + \frac{(y+2)^2}{16} = 1 \] \[ a^2 = 25 \Rightarrow a = 5 \]
Step 2: Check if point lies on ellipse
For $R(-2,-2)$: \[ \frac{(-2-3)^2}{25} + \frac{(-2+2)^2}{16} = \frac{25}{25} + 0 = 1 \] So, $R$ lies on ellipse.
Step 3: Apply property
\[ RS + RT = 2a = 2 \times 5 = 10 \] Final Conclusion:
\[ = 10 \]
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