Maths Q13 – JEE-MAIN 2025

Let

e_{1}

and

e_{2}

be the eccentricities of the ellipse

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

and the hyperbola

\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1

, respectively. If

b < 5

and

e_{1} e_{2} = 1

, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

Clue 1 of 3: Eccentricity Formulas

Recall the formulas for eccentricity of an ellipse and a hyperbola.

🧠 Full Solution Path

Step 1: Calculate Eccentricities and Foci of Given Conics✦ Active

For the ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{25} = 1$ : Since $b < 5$ , the major axis is along the y-axis. $a_{1}^{2} = 25$ , $b_{1}^{2} = b^{2}$ . Eccentricity $e_{1} = \sqrt{1 - \frac{b_{1}^{2}}{a_{1}^{2}}} = \sqrt{1 - \frac{b^{2}}{25}} = \frac{\sqrt{25 - b^{2}}}{5}$ . Its foci are $(0, \pm a_{1} e_{1}) = (0, \pm \sqrt{25 - b^{2}})$ . Let $c_{1} = \sqrt{25 - b^{2}}$ .

For the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1$ : The transverse axis is along the x-axis. $a_{2}^{2} = 16$ , $b_{2}^{2} = b^{2}$ . Eccentricity $e_{2} = \sqrt{1 + \frac{b_{2}^{2}}{a_{2}^{2}}} = \sqrt{1 + \frac{b^{2}}{16}} = \frac{\sqrt{16 + b^{2}}}{4}$ . Its foci are $(\pm a_{2} e_{2}, 0) = (\pm \sqrt{16 + b^{2}}, 0)$ . Let $c_{2} = \sqrt{16 + b^{2}}$ .

Step 2: Determine the value of 'b'○ Expand

Using the condition $e_{1} e_{2} = 1$ :

(\frac{\sqrt{25 - b^{2}}}{5}) (\frac{\sqrt{16 + b^{2}}}{4}) = 1

\sqrt{(25 - b^{2}) (16 + b^{2})} = 20

(25 - b^{2}) (16 + b^{2}) = 400

400 + 25 b^{2} - 16 b^{2} - b^{4} = 400

9 b^{2} - b^{4} = 0

b^{2} (9 - b^{2}) = 0

Since $b$ is a semi-axis length, $b \neq 0$ . Thus, $9 - b^{2} = 0 ⟹ b^{2} = 9 ⟹ b = 3$ (as $b < 5$ ).

💡 Teacher's Secret Hint

Remember to consider the given constraint $b < 5$ when choosing the value of $b$ .

Step 3: Find the Eccentricity of the New Ellipse○ Expand

Substitute $b = 3$ into the foci coordinates:

Foci of ellipse 1: $(0, \pm \sqrt{25 - 3^{2}}) = (0, \pm \sqrt{16}) = (0, \pm 4)$ . These are the y-intercepts of the new ellipse.

Foci of hyperbola 1: $(\pm \sqrt{16 + 3^{2}}, 0) = (\pm \sqrt{25}, 0) = (\pm 5, 0)$ . These are the x-intercepts of the new ellipse.

The new ellipse passes through $(0, \pm 4)$ and $(\pm 5, 0)$ . Since $5 > 4$ , the major axis is along the x-axis. The equation of the new ellipse is $\frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$ , where $A = 5$ and $B = 4$ . So, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ .

The eccentricity $e$ of this new ellipse is given by $e = \sqrt{1 - \frac{B^{2}}{A^{2}}}$ :

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

💡 Teacher's Secret Hint

Ensure you correctly identify the major and minor axes for the new ellipse based on the coordinates of the foci it passes through.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Eccentricity Formulas

🧠 Full Solution Path