1. Standard ellipse and its foci
The standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b>0$ has its major axis along the x-axis. The key parameters are: semi-major axis $a$, semi-minor axis $b$, and the focal distance $c=\sqrt{a^2-b^2}$. The two foci are located at $S(c,0)$ and $S'(-c,0)$. The eccentricity $e=c/a$ satisfies $0
2. Focal distance formulas: $SP = a – ex$ and $S’P = a + ex$
For a point $P(x_0, y_0)$ on the ellipse, the distances to the two foci are given by the elegant formulas $SP = a – ex_0$ and $S’P = a + ex_0$, where $e$ is the eccentricity and $S = (c,0)$, $S’=(-c,0)$. These are derived directly from the definition of an ellipse as the locus of points where the sum of focal distances is constant ($2a$). These formulas avoid the need to compute distances using the Pythagorean theorem, which would involve $\sqrt{(x_0-c)^2+y_0^2}$. The product $SP \cdot S’P = (a-ex_0)(a+ex_0) = a^2 – e^2x_0^2$. This product is what links the given algebraic condition to the coordinates of $P$. Mastering these formulas is essential for all JEE ellipse questions involving foci.
3. Algebraic manipulation of $r_1^2+r_2^2-r_1r_2$ using $r_1+r_2$
The expression $r_1^2+r_2^2-r_1r_2$ appears frequently in conic problems. The standard identity $(r_1+r_2)^2 = r_1^2+r_2^2+2r_1r_2$ lets you write $r_1^2+r_2^2 = (r_1+r_2)^2 – 2r_1r_2$. Substituting: $r_1^2+r_2^2-r_1r_2 = (r_1+r_2)^2 – 3r_1r_2$. If we know $r_1+r_2 = 2a = 10$, then $(r_1+r_2)^2 = 100$. The given condition $100 – 3r_1r_2 = 37$ immediately gives $r_1r_2 = 21$. This three-step manipulation (expand square, substitute, solve) is a powerful technique. Notice also that $r_1^2+r_2^2-r_1r_2 = (r_1-r_2)^2 + r_1r_2$, which is an alternate useful form.
4. Using the ellipse equation to find the second coordinate
Once one coordinate of a point on an ellipse is known, the other is found directly by substituting into the ellipse equation. This is a standard two-step process: (i) find one coordinate through the focal or geometric condition, (ii) substitute into the ellipse equation to get the other. Here, once $\alpha = 5/2$ is found, $\beta^2$ follows immediately: $\frac{(5/2)^2}{25}+\frac{\beta^2}{9}=1$ gives $\beta^2 = 27/4$. Then $\alpha^2+\beta^2 = 25/4 + 27/4 = 13$. Note: since $P$ is in the first quadrant, $\beta > 0$, but we only need $\beta^2$ for the answer. Always use $\beta^2$ directly from the ellipse equation rather than finding $\beta$ and then squaring, to avoid rounding errors.
Q
What are the foci of $\frac{x^2}{25}+\frac{y^2}{9}=1$?
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$c^2=25-9=16$, so $c=4$. Foci are $S(4,0)$ and $S'(-4,0)$.
Q
What is the eccentricity?
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$e = c/a = 4/5$.
Q
What is the formula for $SP$ and $S’P$?
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$SP = a – e\alpha = 5 – \frac{4\alpha}{5}$ and $S’P = a + e\alpha = 5 + \frac{4\alpha}{5}$.
$SP + S’P = 2a = 10$ (fundamental property of the ellipse).
Q
How to find $r_1r_2$ from the given condition?
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$r_1^2+r_2^2-r_1r_2 = (r_1+r_2)^2-3r_1r_2 = 100-3r_1r_2 = 37$, so $r_1r_2=21$.
Q
How to find $\alpha$ from $r_1r_2=21$?
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$r_1r_2 = 25-\frac{16\alpha^2}{25}=21$, so $\alpha^2=25/4$, $\alpha=5/2$.
Substitute $\alpha=5/2$ in ellipse: $1/4+\beta^2/9=1$, so $\beta^2=27/4$.
Q
What is $\alpha^2+\beta^2$?
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$\alpha^2+\beta^2 = 25/4+27/4 = 52/4 = 13$.
Q
Why use focal distance formula instead of coordinate distance?
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The formula $SP=a-e\alpha$ avoids computing $\sqrt{(\alpha-4)^2+\beta^2}$, making algebra much cleaner and faster.
Q
Which option is correct?
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Option (A): $\alpha^2+\beta^2 = 13$.