Find locus of point such that sum of distances from (3,0) and (-3,0) is 10

Question

Find the locus of a point $P(x, y)$ such that the sum of its distances from the points $F_1(3, 0)$ and $F_2(-3, 0)$ is always equal to 10.

Solution — Step by Step

The condition “sum of distances from two fixed points is constant” is the definition of an ellipse. The two fixed points are the foci of the ellipse.

Here: Foci are $F_1(3, 0)$ and $F_2(-3, 0)$ , and the constant sum = 10.

So $PF_1 + PF_2 = 10$ , which means $2a = 10 \Rightarrow a = 5$ (where $a$ is the semi-major axis).

The foci are both on the x-axis, symmetric about the origin:

$c = 3$ (distance from centre to each focus)
$a = 5$ (semi-major axis)

Using the ellipse relation $b^2 = a^2 - c^2$ :

b^2 = 25 - 9 = 16 \Rightarrow b = 4

where $b$ is the semi-minor axis.

The standard form of an ellipse with foci on the x-axis and center at origin:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Substituting $a^2 = 25$ and $b^2 = 16$ :

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

This is the locus. Every point $(x, y)$ on this ellipse satisfies $PF_1 + PF_2 = 10$ .

At the end of the major axis (vertex), $P = (5, 0)$ :

$PF_1 = |5 - 3| = 2$
$PF_2 = |5 - (-3)| = 8$
Sum = $2 + 8 = 10$ ✓

At the end of the minor axis, $P = (0, 4)$ :

$PF_1 = \sqrt{9 + 16} = \sqrt{25} = 5$
$PF_2 = \sqrt{9 + 16} = \sqrt{25} = 5$
Sum = $5 + 5 = 10$ ✓

For an ellipse to exist, the sum of distances must be **greater than** the distance between the foci.

Distance between foci = $|F_1 F_2| = 2c = 6$

Sum of distances = 10 > 6 ✓

If the sum equalled 6, the locus would be the line segment between the foci (a degenerate ellipse). If sum < 6, no locus exists.

Why This Works

The ellipse is geometrically defined as the set of all points where the sum of distances to two fixed points (foci) is constant. Our algebraic derivation uses:

The distance formula to express $PF_1$ and $PF_2$
The relationship $b^2 = a^2 - c^2$ connecting the semi-axes and the focal distance

The equation $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ is the algebraic representation of this geometric locus.

Alternative Method

You can also derive it algebraically without recognising the ellipse:

Let $P = (x, y)$ . Set up $PF_1 + PF_2 = 10$ :

\sqrt{(x-3)^2 + y^2} + \sqrt{(x+3)^2 + y^2} = 10

Move one radical to the right, square both sides, simplify — after two rounds of squaring and simplification, you arrive at $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ . This is more tedious but shows where the standard form comes from.

For JEE Main, this type of problem (“find the locus given a distance condition”) should immediately trigger recognition: sum of distances from two fixed points → ellipse; difference of distances → hyperbola; fixed distance from one point → circle; fixed distance from one point equals fixed distance from a line → parabola. These four identifications save enormous time.

Common Mistake

Students often forget to check whether $2a > 2c$ . If the given sum of distances were less than or equal to 6 (the distance between the two foci), no ellipse exists. Always verify: constant sum > distance between foci, otherwise the problem has no valid locus.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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