Question:medium
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
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