Let the product of the focal distances of the point ( 3, 12) on the ellipse x^2a^2 + y^2b^2 = 1, (a b), be 74. Then the absolute difference of the eccentricities of two such ellipses is:

Given: The point ( 3, 12) lies on the ellipse x^2a^2 + y^2b^2 = 1 with ab, and the product of its focal distances is 74. Find: The absolute difference of the two possible eccentricities. Using the relations shown in the solution: (a + e 3)(a - e 3) = 74 so that a^2 - 3e^2 = 74 Also, since the point lies on the ellipse, 3a^2 + 14b^2 = 1 and b^2 = a^2(1-e^2) Substitute b^2 = a^2(1-e^2) into 3a^2 + 14b^2 = 1 to obtain an equation only in a and e. Combining this with a^2 - 3e^2 = 74 the solution gives 12e^4 - 17e^2 + 6 = 0 Factor the quartic as (3e^2 - 2)(4e^2 - 3) = 0 Hence, e = 23 or e = 34 = 32 Therefore, |e1-e2| = 32 - 23 = 3-2 22 3 Therefore, the correct option is C.

Let the product of the focal distances of the point ( 3, 12) on t… — Solution | JEE Mathematics

MCQMediumJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

Let the product of the focal distances of the point $\left( \sqrt{3}, \frac{1}{2} \right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a > b),$ be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipses is:

A
$\frac{3 - 2\sqrt{2}}{3\sqrt{2}}$
B
$\frac{1 - \sqrt{3}}{\sqrt{2}}$
C
$\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$
D
$\frac{1 - 2\sqrt{2}}{\sqrt{3}}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The point $\left( \sqrt{3}, \frac{1}{2} \right)$ lies on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a>b$ , and the product of its focal distances is $\frac{7}{4}$ .

Find: The absolute difference of the two possible eccentricities.

Using the relations shown in the solution:

(a + e\sqrt{3})(a - e\sqrt{3}) = \frac{7}{4}

so that

a^2 - 3e^2 = \frac{7}{4}

Also, since the point lies on the ellipse,

\frac{3}{a^2} + \frac{1}{4b^2} = 1

and

b^2 = a^2(1-e^2)

Eliminating $$a$$ and solving for $$e$$

Substitute $b^2 = a^2(1-e^2)$ into

\frac{3}{a^2} + \frac{1}{4b^2} = 1

to obtain an equation only in $a$ and $e$ . Combining this with

a^2 - 3e^2 = \frac{7}{4}

the solution gives

12e^4 - 17e^2 + 6 = 0

Factor the quadratic in $$e^2$$

Factor the quartic as

(3e^2 - 2)(4e^2 - 3) = 0

Hence,

e = \sqrt{\frac{2}{3}} \quad \text{or} \quad e = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

Therefore,

|e_1-e_2| = \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} = \frac{3-2\sqrt{2}}{2\sqrt{3}}

Therefore, the correct option is C.

Common mistakes

Using only the point-on-ellipse condition and ignoring the focal-distance product is wrong, because the question requires both conditions to determine the two possible eccentricities. Use both equations together.
Treating $e$ directly as the focal distance parameter $c$ is incorrect. For an ellipse, $c = ae$ and $b^2 = a^2(1-e^2)$ . Keep these quantities distinct while substituting.
Solving $12e^4 - 17e^2 + 6 = 0$ incorrectly by treating it as a quadratic in $e$ instead of $e^2$ leads to wrong roots. First set $u=e^2$ , solve for $u$ , and then take valid positive square roots.

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Answer

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