If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse 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

### Related Formula textSection Formula: quad P = fracm cdot Q + n cdot Om + n$$\text{Section Formula:} \quad P = \frac{m \cdot Q + n \cdot O}{m + n}$$ textChord bisected at (x_1, y_1) : quad T = S_1$$\text{Chord bisected at } (x_1, y_1) : \quad T = S_1$$ ### Core Logic Given parabola x^2 = 4y$x^2 = 4y$, its vertex O = (0, 0)$O = (0, 0)$. A general point Q$Q$ on x^2 = 4y$x^2 = 4y$ is (2t, t^2)$(2t, t^2)$. Let P(h, k)$P(h, k)$ divide OQ$OQ$ in ratio 2:3$2:3$. By section formula: h = frac2(2t) + 3(0)5 = frac4t5$$h = \frac{2(2t) + 3(0)}{5} = \frac{4t}{5}$$ k = frac2(t^2) + 3(0)5 = frac2t^25$$k = \frac{2(t^2) + 3(0)}{5} = \frac{2t^2}{5}$$

### Step 1: Finding the Locus C From h = frac4t5$h = \frac{4t}{5}$, we get t = frac5h4$t = \frac{5h}{4}$. Substitute into k$k$: k = frac25 left(frac5h4right)^2 = frac25 cdot frac25h^216 = frac5h^28$$k = \frac{2}{5} \left(\frac{5h}{4}\right)^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$$ 8k = 5h^2 Rightarrow 5x^2 = 8y$$8k = 5h^2 \Rightarrow 5x^2 = 8y$$ So the conic C is the parabola 5x^2 = 8y$5x^2 = 8y$. ### Step 2: Chord bisected at a point We need the equation of the chord of C: 5x^2 - 8y = 0$C: 5x^2 - 8y = 0$ bisected at (x_1, y_1) = (1, 2)$(x_1, y_1) = (1, 2)$. Use T = S_1$T = S_1$. T = 5xx_1 - 4(y + y_1) = 5x(1) - 4(y + 2) = 5x - 4y - 8$T = 5xx_1 - 4(y + y_1) = 5x(1) - 4(y + 2) = 5x - 4y - 8$ S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$ Equating T$T$ and S_1$S_1$: 5x - 4y - 8 = -11$$5x - 4y - 8 = -11$$ 5x - 4y + 3 = 0$5x - 4y + 3 = 0$ ### Pattern Recognition Internal division locus of a vertex chord on standard conic identically scales the conic. Once the child-conic is found, standard mid-point chord protocol (T=S_1$T=S_1$) strictly applies algebraically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Parabola Class 11 Maths: Straight Lines

### Related Formula textParametric coordinates on y^2 = 4ax: quad (at^2, \, 2at)$$\text{Parametric coordinates on } y^2 = 4ax: \quad (at^2, \, 2at)$$ textFocal Chord relation: t_1 t_2 = -1$$\text{Focal Chord relation: } t_1 t_2 = -1$$ textSection Formula: (x_c, y_c) = left( fracm x_2 + n x_1m+n, \, fracm y_2 + n y_1m+n right)$$\text{Section Formula: } (x_c, y_c) = \left( \frac{m x_2 + n x_1}{m+n}, \, \frac{m y_2 + n y_1}{m+n} \right)$$ ### Core Logic We find the parametric parameters of coordinates P$P$ and Q$Q$, obtain their Cartesian values, and then apply the section formula with the focus S$S$ to calculate the splitting ratio. ### Step 1: Find coordinates of P and Q For parabola y^2 = 16x$y^2 = 16x$, the focal parameter is a = 4$a = 4$. Focus is S(4, 0)$S(4, 0)$. Let P$P$ be (a t_1^2, 2a t_1) = (1, -4)$(a t_1^2, 2a t_1) = (1, -4)$: 2a t_1 = -4 implies 2(4) t_1 = -4 implies t_1 = -frac12$$2a t_1 = -4 \implies 2(4) t_1 = -4 \implies t_1 = -\frac{1}{2}$$ Since PQ$PQ$ is a focal chord, the parametric points are coupled: t_1 t_2 = -1 implies t_2 = 2$$t_1 t_2 = -1 \implies t_2 = 2$$ Now, calculate the coordinates of Q$Q$: Q equiv (a t_2^2, \, 2 a t_2) = (4(4), \, 2(4)(2)) = (16, \, 16)$$Q \equiv (a t_2^2, \, 2 a t_2) = (4(4), \, 2(4)(2)) = (16, \, 16)$$ ### Step 2: Solve for the dividing ratio Let the focus S(4, 0)$S(4, 0)$ divide the line segment PQ$PQ$ internally in the ratio lambda : 1$\lambda : 1$. Using the y$y$-coordinate of the section formula: y_s = fraclambda y_q + 1 y_plambda + 1$$y_s = \frac{\lambda y_q + 1 y_p}{\lambda + 1}$$ 0 = fraclambda(16) + 1(-4)lambda + 1 implies 16lambda - 4 = 0 implies lambda = frac14$$0 = \frac{\lambda(16) + 1(-4)}{\lambda + 1} \implies 16\lambda - 4 = 0 \implies \lambda = \frac{1}{4}$$ Thus, the focus S$S$ divides the chord internally in the ratio 1:4$1:4$. Since gcd(1, 4) = 1$\gcd(1, 4) = 1$, we have m = 1$m = 1$ and n = 4$n = 4$: m^2 + n^2 = 1^2 + 4^2 = 1 + 16 = 17$$m^2 + n^2 = 1^2 + 4^2 = 1 + 16 = 17$$ ### Pattern Recognition Harmonic Mean Shortcut: In any parabola, the focus divides a focal chord internally into segments of lengths SP$SP$ and SQ$SQ$ such that the semi-latus rectum 2a$2a$ is the harmonic mean of these segments: frac1SP + frac1SQ = frac1a$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula For a standard hyperbola: Focus: (pm ae, 0)$(\pm ae, 0)$ Directrix: x = pm fracae$x = \pm \frac{a}{e}$ Eccentricity relation: (ae)^2 = a^2 + b^2$(ae)^2 = a^2 + b^2$ Length of latus rectum: l = frac2b^2a$l = \frac{2b^2}{a}$ ### Core Logic Given ae = sqrt10$ae = \sqrt{10}$ and fracae = frac9sqrt10$\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these gives a^2$a^2$, which determines both parameters. ### Step 1: Find a and e a^2 = (ae) cdot left(fracaeright) = sqrt10 cdot frac9sqrt10 = 9 implies a = 3$$a^2 = (ae) \cdot \left(\frac{a}{e}\right) = \sqrt{10} \cdot \frac{9}{\sqrt{10}} = 9 \implies a = 3$$ Substitute a = 3$a = 3$ into ae = sqrt10$ae = \sqrt{10}$: e = fracsqrt103 implies e^2 = frac109$$e = \frac{\sqrt{10}}{3} \implies e^2 = \frac{10}{9}$$ ### Step 2: Find b and l Using (ae)^2 = a^2 + b^2$(ae)^2 = a^2 + b^2$: 10 = 9 + b^2 implies b^2 = 1$$10 = 9 + b^2 \implies b^2 = 1$$ Then the length of latus rectum l$l$ is: l = frac2b^2a = frac2(1)3 = frac23$$l = \frac{2b^2}{a} = \frac{2(1)}{3} = \frac{2}{3}$$ ### Step 3: Evaluate Final Expression Calculate 9(e^2 + l)$9(e^2 + l)$: 9left(frac109 + frac23right) = 10 + 6 = 16$$9\left(\frac{10}{9} + \frac{2}{3}\right) = 10 + 6 = 16$$ ### Pattern Recognition Multiplying focus location by directrix location immediately eliminates e$e$, giving a^2$a^2$ directly. Once a^2$a^2$ is known, b^2$b^2$ follow seamlessly via (ae)^2 = a^2+b^2$(ae)^2 = a^2+b^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections