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NCERT Exemplar · Q57

Q.If P is a point on the ellipse x216+y225=1\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1 whose foci are S and S′', then PS+PS′=8PS + PS' = 8.

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The key idea is that the given ellipse has its major axis along the yy-axis, so the sum of distances from any point on it to the foci equals the length of the major axis, which is 2a=102a = 10, not 88. The statement in the question is false.


1. Understanding the standard form of an ellipse

An ellipse is defined as the set of all points PP such that the sum of distances to two fixed points (the foci SS and S′S') is constant. That constant is 2a2a, where aa is the semi-major axis length.

The given equation is:

x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 1

Here, the denominator under y2y^2 is larger (25>1625 > 16), so the major axis is vertical (along the yy-axis). For an ellipse of the form x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 with a>ba > b, we have:

  • Semi-major axis: a=25=5a = \sqrt{25} = 5
  • Semi-minor axis: b=16=4b = \sqrt{16} = 4

For any ellipse x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 with a>ba > b, the sum of distances from any point on the ellipse to the two foci is 2a2a.

So here, 2a=2×5=102a = 2 \times 5 = 10.

2. Why the statement PS+PS′=8PS + PS' = 8 is wrong

The problem claims that PS+PS′=8PS + PS' = 8. But from the definition of the ellipse, the sum must be 1010. The number 88 is actually 2b2b (twice the semi-minor axis), which is a common confusion — students sometimes mix up aa and bb when the major axis is vertical. …

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