Let \(e_1\) and \(e_2\) be the eccentricities of the ellipse, \(\frac{x^2}{25}+\frac{y^2}{b^2}=1.... VIDEO
Let \(e_1\) and \(e_2\) be the eccentricities of the ellipse, \(\frac{x^2}{25}+\frac{y^2}{b^2}=1(b<5)\) and the hyperbola, \( \frac{x^2}{16}-\frac{y^2}{\mathrm{~b}^2}=1\) respectively satisfying \(e_1 e_2=1\). If \(\alpha\) and \(\beta\) are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair \((\alpha, \beta)\) is equal to : 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live
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