Let the foci of hyperbola coincide with the foci of the ellipse fracx^236 +fracy^216 = 1$\frac{x^2}{36} +\frac{y^2}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:

### Related Formula textEccentricity of ellipse e_1 = sqrt1 - fracb^2a^2$$\text{Eccentricity of ellipse } e_1 = \sqrt{1 - \frac{b^2}{a^2}}$$ textFoci = (pm ae_1, 0)$$\text{Foci} = (\pm ae_1, 0)$$ textLength of Latus Rectum of hyperbola = frac2b_hyp^2a_hyp$$\text{Length of Latus Rectum of hyperbola} = \frac{2b_{hyp}^2}{a_{hyp}}$$ ### Core Logic For the given ellipse fracx^236 + fracy^216 = 1$\frac{x^2}{36} + \frac{y^2}{16} = 1$: a^2 = 36 Rightarrow a = 6$a^2 = 36 \Rightarrow a = 6$ b^2 = 16$b^2 = 16$ e_1 = sqrt1 - frac1636 = sqrt1 - frac49 = fracsqrt53$$e_1 = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$$ Foci of the ellipse are at (pm ae_1, 0) = left(pm 6 cdot fracsqrt53, 0right) = (pm 2sqrt5, 0)$(\pm ae_1, 0) = \left(\pm 6 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\pm 2\sqrt{5}, 0)$. ### Step 1: Establish Hyperbola Parameters Let the hyperbola be fracx^2p^2 - fracy^2q^2 = 1$\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$. Its foci coincide with the ellipse, so the foci of hyperbola are also (pm 2sqrt5, 0)$(\pm 2\sqrt{5}, 0)$. Let e$e$ be the eccentricity of the hyperbola. We are given e = 5$e = 5$. Focus of hyperbola is pe = 2sqrt5$pe = 2\sqrt{5}$. p(5) = 2sqrt5 Rightarrow p = frac2sqrt55 = frac2sqrt5$$p(5) = 2\sqrt{5} \Rightarrow p = \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$$ ### Step 2: Find the Conjugate Axis (q) For the hyperbola: e^2 = 1 + fracq^2p^2$$e^2 = 1 + \frac{q^2}{p^2}$$ 25 = 1 + fracq^2left(frac2sqrt5right)^2$$25 = 1 + \frac{q^2}{\left(\frac{2}{\sqrt{5}}\right)^2}$$ 24 = fracq^24/5 Rightarrow 24 = frac5q^24$$24 = \frac{q^2}{4/5} \Rightarrow 24 = \frac{5q^2}{4}$$ 5q^2 = 96 Rightarrow q^2 = frac965$$5q^2 = 96 \Rightarrow q^2 = \frac{96}{5}$$ ### Step 3: Calculate Latus Rectum Length of Latus Rectum = frac2q^2p$\frac{2q^2}{p}$ = frac2 left(frac965right)frac2sqrt5 = frac965 times sqrt5 = frac96sqrt5$$= \frac{2 \left(\frac{96}{5}\right)}{\frac{2}{\sqrt{5}}} = \frac{96}{5} \times \sqrt{5} = \frac{96}{\sqrt{5}}$$ ### Pattern Recognition Co-focal conics share the exact mathematical value of their focal length ae$ae$ (or pe$pe$). Instantly extract c = ae$c = ae$ from the first shape and map it directly to c = pe$c = pe$ for the second. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections