Let S and S' be the foci of the ellipse x²/25 + y²/9 = 1 and P(α, β) be a point on the ellipse in th

Q: Why do we use $SP + S'P = 2a$ instead of just plugging the coordinates into the ellipse equation?

While you could substitute $(\alpha, \beta)$ into the ellipse equation, it would lead to a messy system of equations involving $\alpha^2$ and $\beta^2$ that is much harder to solve. The focal property is a geometric shortcut that simplifies the algebra significantly.

Q: How do I know which distance is $SP$ and which is $S'P$?

The problem states $P$ is in the first quadrant. Since $S(4, 0)$ is on the positive x-axis and $S'(-4, 0)$ is on the negative x-axis, a point in the first quadrant is physically closer to $S$ than to $S'$. Thus, $SP < S'P$.

Q: Is there a trick to solving the quadratic $t^2 - 10t + 21 = 0$?

Not really a trick, but a standard technique! Whenever you know the sum and product of two numbers, they are always the roots of the quadratic equation $t^2 - (sum)t + (product) = 0$. It's a powerful tool to keep in your arsenal.

Question

Let $S$ and $S'$ be the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $P(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2 + (S'P)^2 - SP \cdot S'P = 37$, then $\alpha^2 + \beta^2$ is equal to :

Options:

(A) 15

(B) 11

(C) 17

(D) 13

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a canvas, tracing the path of a planet orbiting a star. That path is an ellipse, a shape of profound mathematical beauty. We are given the ellipse:

$$
\frac{x^2}{25} + \frac{y^2}{9} = 1
$$

By comparing this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we immediately see that $a^2 = 25$ and $b^2 = 9$. This gives us $a = 5$ and $b = 3$.

The distance from the center to the foci, $c$, is governed by the relationship $c^2 = a^2 - b^2$. Plugging in our values, $c^2 = 25 - 9 = 16$, which means $c = 4$. Thus, our foci are anchored at $S(4, 0)$ and $S'(-4, 0)$.

### The Focal Magic

Now, let us introduce a point $P(\alpha, \beta)$ on this curve. There is a fundamental property of ellipses that every JEE aspirant must hold dear: the sum of the focal distances of any point on the ellipse is constant and equal to the length of the major axis, $2a$.

$$
SP + S'P = 2a = 10
$$

This is our first major breakthrough. It transforms a complex geometric problem into a manageable algebraic one.

### The Algebraic Dance

The problem presents us with the condition: $(SP)^2 + (S'P)^2 - SP \cdot S'P = 37$. Let $r_1 = SP$ and $r_2 = S'P$. Our condition becomes:

$$
r_1^2 + r_2^2 - r_1 r_2 = 37
$$

We know $r_1 + r_2 = 10$. Using the identity $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2$, we substitute this into our condition:

$$
(r_1 + r_2)^2 - 2r_1 r_2 - r_1 r_2 = 37 \Rightarrow (r_1 + r_2)^2 - 3r_1 r_2 = 37
$$

Since $r_1 + r_2 = 10$, we have $100 - 3r_1 r_2 = 37$. Solving for the product, $3r_1 r_2 = 63$, so $r_1 r_2 = 21$.

### Bridging Geometry and Algebra

We now have the sum $r_1 + r_2 = 10$ and the product $r_1 r_2 = 21$. These are the roots of the quadratic equation $t^2 - 10t + 21 = 0$. Factoring this, we get $(t - 3)(t - 7) = 0$.

Thus, the distances are $3$ and $7$. Since $P$ is in the first quadrant, it is closer to $S(4, 0)$ than to $S'(-4, 0)$, so $SP = 3$ and $S'P = 7$. We apply the distance formula:

$$
SP^2 = (\alpha - 4)^2 + \beta^2 = 3^2 = 9
$$

$$
S'P^2 = (\alpha + 4)^2 + \beta^2 = 7^2 = 49
$$

### The Final Calculation

Subtracting the first equation from the second, the $\beta^2$ terms cancel out:

$$
((\alpha + 4)^2 + \beta^2) - ((\alpha - 4)^2 + \beta^2) = 49 - 9
$$

$$
(\alpha^2 + 8\alpha + 16) - (\alpha^2 - 8\alpha + 16) = 40 \Rightarrow 16\alpha = 40 \Rightarrow \alpha = \frac{5}{2}
$$

Substituting $\alpha = \frac{5}{2}$ back into $(\alpha - 4)^2 + \beta^2 = 9$:

$$
(\frac{5}{2} - 4)^2 + \beta^2 = 9 \Rightarrow (-\frac{3}{2})^2 + \beta^2 = 9 \Rightarrow \frac{9}{4} + \beta^2 = 9
$$

$$
\beta^2 = 9 - \frac{9}{4} = \frac{27}{4}
$$

Finally, we calculate $\alpha^2 + \beta^2$:

$$
\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 + \frac{27}{4} = \frac{25}{4} + \frac{27}{4} = \frac{52}{4} = 13
$$

We have arrived at the answer: **13**.

Select Answer:

Visualized Solution

Equation of the Ellipse

Finding the Foci

The Focal Property of Ellipse

The Given Condition

Algebraic Manipulation

Calculating $r_{1} r_{2}$

Finding $S P$ and $S^{'} P$

Applying Distance Formula

Solving for $α$

Solving for $β^{2}$

Final Answer

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

Analyzing the Setup

The Focal Magic

The Algebraic Dance

Bridging Geometry and Algebra

The Final Calculation

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