Let P(10, 2sqrt15)$P(10, 2\sqrt{15})$ be a point on the hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S$S$ and S'$S'$. If the length of its latus rectum is 8$8$, then the square of the area of Delta PSS'$\Delta PSS'$ is equal to:

### 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 textCondition of tangency for ellipse fracx^2a^2 + fracy^2b^2 = 1 text is c^2 = a^2m^2 + b^2$$\text{Condition of tangency for ellipse } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ is } c^2 = a^2m^2 + b^2$$ textFocal distance SP = a pm ex$$\text{Focal distance } SP = a \pm ex$$ textEccentricity e = sqrt1 - fracb^2a^2$$\text{Eccentricity } e = \sqrt{1 - \frac{b^2}{a^2}}$$ ### Core Logic

Identify the slope and intercepts of the line to find alpha$\alpha$. Use the point of contact formula to locate P(x_1, y_1)$P(x_1, y_1)$ and apply focal distance definitions. ### Step 1: Determine alpha Rewrite the ellipse: fracx^21/3 + fracy^21/4 = 1 implies a^2 = frac13, b^2 = frac14$\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1 \implies a^2 = \frac{1}{3}, b^2 = \frac{1}{4}$. The line is y = -fracalpha4x + fracsqrt74$y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$. Using c^2 = a^2m^2 + b^2$c^2 = a^2m^2 + b^2$: left(fracsqrt74right)^2 = frac13 left(-fracalpha4right)^2 + frac14$$\left(\frac{\sqrt{7}}{4}\right)^2 = \frac{1}{3} \left(-\frac{\alpha}{4}\right)^2 + \frac{1}{4}$$ frac716 = fracalpha^248 + frac416 implies frac316 = fracalpha^248 implies alpha^2 = 9 implies alpha = pm 3$$\frac{7}{16} = \frac{\alpha^2}{48} + \frac{4}{16} \implies \frac{3}{16} = \frac{\alpha^2}{48} \implies \alpha^2 = 9 \implies \alpha = \pm 3$$ Since P$P$ is in the first quadrant, coordinates x, y$x, y$ are positive, so we use the tangent 3x + 4y - sqrt7 = 0$3x + 4y - \sqrt{7} = 0$. ### Step 2: Find Point of Contact P The tangent at P(x_1, y_1)$P(x_1, y_1)$ is 3xx_1 + 4yy_1 = 1$3xx_1 + 4yy_1 = 1$. Comparing this with 3x + 4y = sqrt7$3x + 4y = \sqrt{7}$ (divided by sqrt7$\sqrt{7}$ to match constant 1$1$): frac3xsqrt7 + frac4ysqrt7 = 1$\frac{3x}{\sqrt{7}} + \frac{4y}{\sqrt{7}} = 1$. Comparing coefficients: 3x_1 = frac3sqrt7 implies x_1 = frac1sqrt7$$3x_1 = \frac{3}{\sqrt{7}} \implies x_1 = \frac{1}{\sqrt{7}}$$ 4y_1 = frac4sqrt7 implies y_1 = frac1sqrt7$$4y_1 = \frac{4}{\sqrt{7}} \implies y_1 = \frac{1}{\sqrt{7}}$$ So P = left(frac1sqrt7, frac1sqrt7right)$P = \left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$. ### Step 3: Calculate Focal Distance Find eccentricity: e = sqrt1 - frac1/41/3 = sqrt1 - frac34 = frac12$$e = \sqrt{1 - \frac{1/4}{1/3}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$$ The focal distances are a - ex$a - ex$ and a + ex$a + ex$. Since a^2 = 1/3 implies a = 1/sqrt3$a^2 = 1/3 \implies a = 1/\sqrt{3}$. SP = a - ex_1 = frac1sqrt3 - frac12left(frac1sqrt7right) = frac1sqrt3 - frac12sqrt7$$SP = a - ex_1 = \frac{1}{\sqrt{3}} - \frac{1}{2}\left(\frac{1}{\sqrt{7}}\right) = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{7}}$$ S'P = a + ex_1 = frac1sqrt3 + frac12left(frac1sqrt7right) = frac1sqrt3 + frac12sqrt7$$S'P = a + ex_1 = \frac{1}{\sqrt{3}} + \frac{1}{2}\left(\frac{1}{\sqrt{7}}\right) = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$$ Matching with the options, the focal distance is frac1sqrt3 + frac12sqrt7$\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$. ### Pattern Recognition For tangency lx+my+n=0$lx+my+n=0$ to x^2/a^2 + y^2/b^2 = 1$x^2/a^2 + y^2/b^2 = 1$, use a^2 l^2 + b^2 m^2 = n^2$a^2 l^2 + b^2 m^2 = n^2$. Points of contact can be quickly evaluated by comparing T=0$T=0$ to the normalized tangent equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

### Related Formula textCentroid G(x,y) = left( fracx_1 + x_2 + x_33, fracy_1 + y_2 + y_33 right)$$\text{Centroid } G(x,y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$$ textProduct of slopes for perpendicular lines m_1 m_2 = -1$$\text{Product of slopes for perpendicular lines } m_1 m_2 = -1$$ ### Core Logic

Define coordinates: O(0,0)$O(0,0)$, P(3t^2, 6t)$P(3t^2, 6t)$ (since y^2 = 12x implies 4a = 12 implies a=3$y^2 = 12x \implies 4a = 12 \implies a=3$). Determine the slope of OP$OP$ and use perpendicularity to find the equation of PA$PA$ and locate point A$A$ on the x-axis. ### Step 1: Locate A via Perpendicularity Slope of OP$OP$ is m_OP = frac6t - 03t^2 - 0 = frac2t$m_{OP} = \frac{6t - 0}{3t^2 - 0} = \frac{2}{t}$. Since angle OPA = 90^circ$\angle OPA = 90^\circ$, slope of AP$AP$ is m_AP = -fract2$m_{AP} = -\frac{t}{2}$. Equation of AP$AP$: y - 6t = -fract2(x - 3t^2)$$y - 6t = -\frac{t}{2}(x - 3t^2)$$ To find A$A$ on the x-axis, put y = 0$y = 0$: -6t = -fract2(x - 3t^2) implies 12 = x - 3t^2 implies x = 12 + 3t^2$$-6t = -\frac{t}{2}(x - 3t^2) \implies 12 = x - 3t^2 \implies x = 12 + 3t^2$$ So, A$A$ is (12 + 3t^2, 0)$(12 + 3t^2, 0)$. ### Step 2: Locus of the Centroid Let the centroid of triangle OPA$\triangle OPA$ be G(h, k)$G(h, k)$. h = frac0 + 3t^2 + (12 + 3t^2)3 = frac6t^2 + 123 = 2t^2 + 4$$h = \frac{0 + 3t^2 + (12 + 3t^2)}{3} = \frac{6t^2 + 12}{3} = 2t^2 + 4$$ k = frac0 + 6t + 03 = 2t$$k = \frac{0 + 6t + 0}{3} = 2t$$ From k = 2t implies t = frack2$k = 2t \implies t = \frac{k}{2}$. Substitute t$t$ into the equation for h$h$: h = 2left(frack2right)^2 + 4 = frack^22 + 4$$h = 2\left(\frac{k}{2}\right)^2 + 4 = \frac{k^2}{2} + 4$$ 2h = k^2 + 8 implies k^2 = 2h - 8$$2h = k^2 + 8 \implies k^2 = 2h - 8$$ Replacing (h, k)$(h, k)$ with (x, y)$(x, y)$, the locus is y^2 = 2x - 8 implies y^2 - 2x + 8 = 0$y^2 = 2x - 8 \implies y^2 - 2x + 8 = 0$. ### Pattern Recognition For any right-angled configuration involving the origin and axis on a parabola, parametric geometry simplifies equations significantly. Find coordinates O, P, A$O, P, A$, set up the centroid algebraic relations, and eliminate parameter t$t$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

### Related Formula textFor a focal chord with ends (at_1^2, 2at_1) text and (at_2^2, 2at_2), text the relation is t_1t_2 = -1$$\text{For a focal chord with ends } (at_1^2, 2at_1) \text{ and } (at_2^2, 2at_2), \text{ the relation is } t_1t_2 = -1$$ textSection formula: (x, y) = left( fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+n right)$$\text{Section formula: } (x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$$ ### Core Logic

For y^2 = 16x$y^2 = 16x$, a = 4$a = 4$. The given point A(16, 16)$A(16, 16)$ is equivalent to 4t^2 = 16$4t^2 = 16$ and 2(4)t = 16$2(4)t = 16$, which gives parameter t_1 = 2$t_1 = 2$. The other end B$B$ has parameter t_2 = -frac1t_1 = -frac12$t_2 = -\frac{1}{t_1} = -\frac{1}{2}$. ### Step 1: Calculate coordinates of B For t_2 = -1/2$t_2 = -1/2$, point B$B$ is: x = 4(-1/2)^2 = 1$x = 4(-1/2)^2 = 1$ y = 8(-1/2) = -4$y = 8(-1/2) = -4$ So, B(1, -4)$B(1, -4)$. ### Step 2: Section formula calculations (Two cases) Point P(alpha, beta)$P(\alpha, \beta)$ divides AB$AB$ in the ratio 5:2$5:2$. There are two possibilities depending on which end the ratio starts from. Case 1: Ratio 5 from B to A (i.e. A is x_2$x_2$ and B is x_1$x_1$): alpha = frac5(16) + 2(1)7 = frac80 + 27 = frac827$$\alpha = \frac{5(16) + 2(1)}{7} = \frac{80 + 2}{7} = \frac{82}{7}$$ beta = frac5(16) + 2(-4)7 = frac80 - 87 = frac727$$\beta = \frac{5(16) + 2(-4)}{7} = \frac{80 - 8}{7} = \frac{72}{7}$$ Sum: alpha + beta = frac1547 = 22$\alpha + \beta = \frac{154}{7} = 22$. Case 2: Ratio 5 from A to B (i.e. B is x_2$x_2$ and A is x_1$x_1$): alpha = frac5(1) + 2(16)7 = frac5 + 327 = frac377$$\alpha = \frac{5(1) + 2(16)}{7} = \frac{5 + 32}{7} = \frac{37}{7}$$ beta = frac5(-4) + 2(16)7 = frac-20 + 327 = frac127$$\beta = \frac{5(-4) + 2(16)}{7} = \frac{-20 + 32}{7} = \frac{12}{7}$$ Sum: alpha + beta = frac497 = 7$\alpha + \beta = \frac{49}{7} = 7$. ### Step 3: Minimum Value Comparing the two possible sums, 7 < 22$7 < 22$. Thus, the minimum value is 7$7$. ### Pattern Recognition When a line segment is divided in a given ratio, 'internal division' inherently bears two solutions based on the orientation (from point A or point B). Always evaluate both cases when finding a minimum or maximum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections