If the co-ordinates of two points A and B are (√7, 0) and (-√7, 0) respectively and P is any point o

Q: Wait, why can't I just use the distance formula? It seems straightforward to calculate $\sqrt{(x-\sqrt{7})^2 + y^2} + \sqrt{(x+\sqrt{7})^2 + y^2}$.

Ah, the trap of brute force! While technically possible, it is an algebraic nightmare that will consume your time and sanity. JEE Advanced tests your conceptual intuition, not your stamina for messy algebra. By recognizing the focal property, you bypass pages of calculation and arrive at the answer in seconds.

Q: How do I know if the major axis is along the $x$-axis or $y$-axis?

Look at the denominators in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If the denominator under $x^2$ is larger, the ellipse is stretched horizontally. If the denominator under $y^2$ is larger, it's vertical. Here, $16 > 9$, so it's horizontal.

Q: Why is the sum of distances always $2a$?

Imagine pinning a string of length $2a$ at the two foci and pulling it taut with a pencil. The path traced by the pencil is the ellipse. By definition, the string length is the sum of the distances from the pencil to the two pins. That string length is the major axis, $2a$.

Question

If the co-ordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic, $9x^2 + 16y^2 = 144$, then $PA + PB$ is equal to :

Options:

(A) 9

(B) 16

(C) 6

(D) 8

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unlocking the Ellipse

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey into the heart of conic sections.

Often, when we see an equation like $9x^2 + 16y^2 = 144$, our instinct is to reach for the distance formula, to grind through the algebra, and to hope for the best. But in the world of JEE Advanced, the most powerful tool in your arsenal is not your calculator—it is your ability to see the hidden geometry behind the numbers.

### Phase 1: The Transformation

Let us look at our equation: $9x^2 + 16y^2 = 144$. It looks a bit cluttered, doesn't it? To understand its soul, we must bring it into its standard form.

We divide the entire equation by $144$, yielding:

$$\frac{9x^2}{144} + \frac{16y^2}{144} = \frac{144}{144}$$

This simplifies beautifully to:

$$\frac{x^2}{16} + \frac{y^2}{9} = 1$$

Now, the ellipse reveals its true nature. We compare this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We immediately see that $a^2 = 16$ and $b^2 = 9$.

This tells us that $a = 4$ and $b = 3$. Because $a > b$, we know our ellipse is stretched horizontally along the $x$-axis. We have our parameters; now, let us find the foci.

### Phase 2: The Revelation

To find the foci, we need the eccentricity, $e$. The formula is $e = \sqrt{1 - \frac{b^2}{a^2}}$.

Substituting our values, we get:

$$e = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$

Now, the coordinates of the foci are given by $(\pm ae, 0)$. Let us calculate $ae$:

$$ae = 4 	imes \frac{\sqrt{7}}{4} = \sqrt{7}$$

So, the foci are at $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$.

Stop for a moment. Look at the points $A$ and $B$ given in the problem statement. They are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$. They are the foci!

This is the "Aha!" moment. The problem wasn't asking us to calculate a random distance; it was testing our understanding of the definition of an ellipse.

### Phase 3: The Definition

What is an ellipse? It is the locus of all points $P$ such that the sum of the distances from two fixed points (the foci) is constant. And what is that constant? It is the length of the major axis, which is $2a$.

We have already determined that $a = 4$. Therefore, for any point $P$ on this ellipse, the sum of the distances $PA + PB$ must be equal to $2a$.

$$PA + PB = 2(4) = 8$$

### Conclusion

There is a profound lesson here. If you had tried to use the distance formula, you would have been trapped in a labyrinth of square roots and algebraic expansion.

But by pausing to identify the geometric properties, the solution collapsed into a single, elegant step. This is the essence of JEE Advanced physics and mathematics: look for the symmetry, respect the definitions, and let the geometry guide you to the answer.

You have mastered the ellipse today. The final answer is **8**. Keep that clarity of thought, and you will conquer any problem that comes your way.

Animated Solution for Mathematics - Conic Sections: If the co-ordinates of two points $A$ and $B$ are $(7, 0)$ and $(- 7, 0)$ respectively and $P$ is any point on the conic, $9 x^{2} + 16 y^{2} = 144$ , then $P A + P B$ is equal to :

Select Answer:

Visualized Solution

Given Points $A$ and $B$

The Conic Equation

Standard Form of Ellipse

Extracting $a$ and $b$

Eccentricity Formula

Calculating Eccentricity

Coordinates of Foci

The Revelation

Point $P$ on the Ellipse

The Focal Distance Property

Final Answer

The Sigma Insight: Foci, Directrices, and Eccentricity

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

If $S$ and $S^{'}$ are the foci of the ellipse $\frac{x ^{2}}{18} + \frac{y ^{2}}{9} = 1$ and $P$ be a point on the ellipse, then $min (S P \cdot S^{'} P) + max (S P \cdot S^{'} P)$ is equal to :

Let $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$ and $a = f \circ g (10), b = g \circ f (3)$ . If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x ^{2}}{a} + \frac{y ^{2}}{b} = 1$ , then $8 e^{2} + l^{2}$ is equal to.

The radius of the circle passing through the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ , and having its centre at $(0, 3)$ is

Let the product of the focal distances of the point $P (4, 23)$ on the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be 32. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$ . Then $p^{2} + q^{2}$ is equal to _______

The equation of the circle passing through the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ , and having centre at $(0, 3)$ is

If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse, $\frac{x ^{2}}{18} + \frac{y ^{2}}{4} = 1$ and the hyperbola, $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse, $15 x^{2} + 3 y^{2} = k$ . Then $k$ is equal to:

Let one focus of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be at $(10, 0)$ and the corresponding directrix be $x = \frac{9}{10}$ . If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$ , then $9 (e^{2} + l)$ is equal to:

If $e_{1}$ and $e_{2}$ are eccentricities of the ellipse $\frac{x ^{2}}{18} + \frac{y ^{2}}{4} = 1$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ respectively and if the point $(e_{1}, e_{2})$ lies on ellipse $15 x^{2} + 3 y^{2} = k$ . Then find value of $k$

If the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ meets the line $\frac{x}{7} + \frac{y}{2 6} = 1$ on the x-axis and the line $\frac{x}{7} - \frac{y}{2 6} = 1$ on the y-axis, then the eccentricity of the ellipse is

Let $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, a > 0, b > 0$ , be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (22 + 14)$ . If the eccentricity $H$ is $\frac{11}{2}$ , then value of $a^{2} + b^{2}$ is equal to ______.