If P = (x, y), F₁ = (3, 0), F₂ = (-3, 0) and 16x² + 25y² = 400, then PF₁ + PF₂ equals (a) 8 (b) 6 (c

Q: Why can't I just use the distance formula $\sqrt{(x-3)^2 + y^2} + \sqrt{(x+3)^2 + y^2}$? It seems more direct.

While mathematically correct, it is a trap! The algebra will become a nightmare of radicals. JEE rewards those who recognize the geometric definition: the sum of focal distances is always $2a$. Always look for the geometric shortcut first.

Q: How do I know for sure that $F_1$ and $F_2$ are the foci? Could they just be random points?

Great question! You must verify. Calculate $ae$ using the parameters $a$ and $b$ from your equation. If $ae$ matches the x-coordinate of your given points, they are indeed the foci. In this problem, $a=5$ and $e=3/5$, so $ae=3$, which perfectly matches the given points.

Q: What if the equation was $25x^2 + 16y^2 = 400$? Would the answer change?

Yes! The major axis would shift to the y-axis. You would need to re-evaluate $a$ and $b$ and the position of the foci. Always check which denominator is larger to determine the orientation of the ellipse.

Question

If $P = (x, y)$, $F_1 = (3, 0)$, $F_2 = (-3, 0)$ and $16x^2 + 25y^2 = 400$, then $PF_1 + PF_2$ equals

Options:

(A) 8

(B) 6

(C) 10

(D) 12

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unveiling the Ellipse

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric truth.

When you look at an equation like $16x^2 + 25y^2 = 400$, do not see it as a cold, rigid set of numbers. See it as a hidden shape waiting to be revealed.

Many students rush into the distance formula, trying to calculate the distance from point $P(x, y)$ to $F_1(3, 0)$ and $F_2(-3, 0)$ using the brute force of algebra. But I want you to pause.

In coordinate geometry, the most elegant path is rarely the one that requires the most ink. Let us embark on a journey to find the soul of this curve.

### Phase 1: The Transformation

Our first step is to bring this equation into the light. The standard form of an ellipse,

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

is our guiding star. To reach it, we must make the right-hand side equal to $1$.

By dividing the entire equation $16x^2 + 25y^2 = 400$ by $400$, we get:

$$ \frac{16x^2}{400} + \frac{25y^2}{400} = \frac{400}{400} $$

Simplifying these fractions is where the magic happens. $16$ goes into $400$ exactly $25$ times, and $25$ goes into $400$ exactly $16$ times.

Our equation transforms into:

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

Now, the ellipse is laid bare. We can clearly see that $a^2 = 25$, which means $a = 5$, and $b^2 = 16$, which means $b = 4$.

Since $a > b$, we know our ellipse is stretched horizontally along the $x$-axis.

### Phase 2: The Detective Work

Now, let us address the points $F_1(3, 0)$ and $F_2(-3, 0)$. Are they just random points? Or are they the keys to the kingdom?

To find out, we calculate the eccentricity $e$ of our ellipse. The formula is:

$$ e = \sqrt{1 - \frac{b^2}{a^2}} $$

Substituting our values, we get:

$$ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $$

The foci of a horizontal ellipse are located at $(\pm ae, 0)$. Let us calculate $ae$:

$$ 5 	imes \frac{3}{5} = 3 $$

The foci are at $(3, 0)$ and $(-3, 0)$. The realization hits: the points given in the problem are exactly the foci of our ellipse!

### Phase 3: The Elegant Conclusion

We have arrived at the heart of the matter. The fundamental definition of an ellipse is the locus of all points such that the sum of the distances from two fixed points (the foci) is constant.

This constant sum is equal to the length of the major axis, which is $2a$. We have already found $a = 5$.

Therefore, the sum $PF_1 + PF_2$ must be:

$$ 2 	imes 5 = 10 $$

We did not need to perform a single complex distance calculation. By recognizing the geometric properties, we bypassed the algebra and arrived at the truth.

The answer is **10**. Remember, in JEE, the most powerful tool in your arsenal is not your calculator, but your ability to see the geometry behind the equations.

Animated Solution for Mathematics - Conic Sections: If $P = (x, y)$ , $F_{1} = (3, 0)$ , $F_{2} = (- 3, 0)$ and $16 x^{2} + 25 y^{2} = 400$ , then $P F_{1} + P F_{2}$ equals

Select Answer:

Visualized Solution

Given Equation

Standardizing the Equation

Dividing by $400$

Simplifying the Equation

Identifying Semi-Axes

Eccentricity Formula

Substituting Values for $e$

Calculating Eccentricity

Locating the Foci

The Focal Property of an Ellipse

Applying the Property

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P=(x,y), F1​=(3,0), F2​=(−3,0) and 16x2+25y2=400, then PF1​+PF2​ equals