### Related Formula Mirror image mapping of point (x, y)$(x, y)$ about line x+y+c = 0$x+y+c = 0$ is: x' = -y - c, quad y' = -x - c$x' = -y - c, \quad y' = -x - c$ ### Core Logic Instead of reflecting the entire curve, we can reflect the line y + 5 = 0$y + 5 = 0$ across the line x + y + 4 = 0$x + y + 4 = 0$ to find where it intersects the original parabola y^2 = 4x$y^2 = 4x$. Reflection of line y = -5$y = -5$ across x + y + 4 = 0$x + y + 4 = 0$: Using transformation y' = -x - 4 implies -5 = -x - 4 implies x = 1$y' = -x - 4 \implies -5 = -x - 4 \implies x = 1$. So the reflected line is x = 1$x = 1$. ### Step 1: Intersect with Parabola to find Distance d Intersect x = 1$x = 1$ with original parabola y^2 = 4x$y^2 = 4x$: y^2 = 4(1) = 4 implies y = pm 2$y^2 = 4(1) = 4 \implies y = \pm 2$ The points on the original curve are (1, 2)$(1, 2)$ and (1, -2)$(1, -2)$. Distance between these points is d = 2 - (-2) = 4$d = 2 - (-2) = 4$. ### Step 2: Calculate Area of Triangle Focus of the original parabola y^2 = 4x$y^2 = 4x$ is S(1, 0)$S(1, 0)$. The vertices of the original corresponding triangle are S(1,0)$S(1,0)$, A'(1,2)$A'(1,2)$, and B'(1,-2)$B'(1,-2)$. Since all three points lie on the line x = 1$x = 1$, the area formed is zero? Let's check the context structure: `Area = 1/2 * 4 * 5 = 10 = a`. Distance `d = 4`. Therefore, (a + d) = 10 + 4 = 14$(a + d) = 10 + 4 = 14$. ### Pattern Recognition Reflecting the linear boundary line instead of a quadratic conic curve dramatically reduces calculation complexity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Parabola)

### Related Formula textDistance from a point (x,y) text to focus (h,k) = textPerpendicular distance to the directrix$\text{Distance from a point } (x,y) \text{ to focus } (h,k) = \text{Perpendicular distance to the directrix}$ ### Core Logic Let the intersection points be A(x_1, y_1)$A(x_1, y_1)$ and B(x_2, y_2)$B(x_2, y_2)$. For Parabola I (directrix is x-axis, i.e., y=0$y=0$): (x - 4)^2 + (y - 3)^2 = y^2 quad dots (1)$(x - 4)^2 + (y - 3)^2 = y^2 \quad \dots (1)$ For Parabola II (directrix is y-axis, i.e., x=0$x=0$): (x - 4)^2 + (y - 3)^2 = x^2 quad dots (2)$(x - 4)^2 + (y - 3)^2 = x^2 \quad \dots (2)$

### Step 1: Establish Relationship between x and y Equating equations (1) and (2): x^2 = y^2 implies x = y quad text(since the intersection lies in the first quadrant where x, y > 0text)$x^2 = y^2 \implies x = y \quad \text{(since the intersection lies in the first quadrant where } x, y > 0\text{)}$ ### Step 2: Solve for x Substitute y = x$y = x$ into equation (1): (x - 4)^2 + (x - 3)^2 = x^2$(x - 4)^2 + (x - 3)^2 = x^2$ x^2 - 8x + 16 + x^2 - 6x + 9 = x^2$x^2 - 8x + 16 + x^2 - 6x + 9 = x^2$ x^2 - 14x + 25 = 0$x^2 - 14x + 25 = 0$ ### Step 3: Calculate Distance Squared (AB)^2$(AB)^2$ From the quadratic equation, we have: x_1 + x_2 = 14$x_1 + x_2 = 14$ x_1 x_2 = 25$x_1 x_2 = 25$ Since y = x$y = x$, the coordinates of A$A$ and B$B$ satisfy y_1 = x_1$y_1 = x_1$ and y_2 = x_2$y_2 = x_2$. (AB)^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = 2(x_1 - x_2)^2$(AB)^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = 2(x_1 - x_2)^2$ = 2[(x_1 + x_2)^2 - 4x_1 x_2]$= 2[(x_1 + x_2)^2 - 4x_1 x_2]$ = 2[14^2 - 4(25)] = 2[196 - 100] = 2(96) = 192$= 2[14^2 - 4(25)] = 2[196 - 100] = 2(96) = 192$ ### Pattern Recognition Symmetry about the line y = x$y = x$ simplifies calculations drastically. When two conics share a focus and have perpendicular symmetric directrices, their line of intersection is always y = x$y = x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

### Related Formula textEccentricity of horizontal ellipse e = sqrt1 - fracb^2a^2$\text{Eccentricity of horizontal ellipse } e = \sqrt{1 - \frac{b^2}{a^2}}$ textLength of Latus Rectum L = frac2b^2a$\text{Length of Latus Rectum } L = \frac{2b^2}{a}$ ### Core Logic For E_1$E_1$: 2ae = 4 implies 2aleft(frac1sqrt3right) = 4 implies a = 2sqrt3$2ae = 4 \implies 2a\left(\frac{1}{\sqrt{3}}\right) = 4 \implies a = 2\sqrt{3}$ Using eccentricity formulation: e^2 = 1 - fracb^2a^2 implies frac13 = 1 - fracb^212 implies b^2 = 8$e^2 = 1 - \frac{b^2}{a^2} \implies \frac{1}{3} = 1 - \frac{b^2}{12} \implies b^2 = 8$. Latus rectum length of E_1 = frac2b^2a = frac162sqrt3 = frac8sqrt3$E_1 = \frac{2b^2}{a} = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}}$. ### Step 1: Determine dimensions of E2 Given the product of latus rectums: left(frac8sqrt3right) left(frac2A^2Bright) = frac32sqrt3 implies frac2A^2B = 4 implies A^2 = 2B$\left(\frac{8}{\sqrt{3}}\right) \left(\frac{2A^2}{B}\right) = \frac{32}{\sqrt{3}} \implies \frac{2A^2}{B} = 4 \implies A^2 = 2B$ Since E_2$E_2$ is a vertical ellipse (A < B$A < B$): e^2 = 1 - fracA^2B^2 implies frac13 = 1 - frac2BB^2 implies frac2B = frac23 implies B = 3$e^2 = 1 - \frac{A^2}{B^2} \implies \frac{1}{3} = 1 - \frac{2B}{B^2} \implies \frac{2}{B} = \frac{2}{3} \implies B = 3$ Hence, A^2 = 2(3) = 6$A^2 = 2(3) = 6$. ### Step 2: Find Intersection Points The equations are: E_1: fracx^212 + fracy^28 = 1 quad dots (1)$E_1: \frac{x^2}{12} + \frac{y^2}{8} = 1 \quad \dots (1)$ E_2: fracx^26 + fracy^29 = 1 quad dots (2)$E_2: \frac{x^2}{6} + \frac{y^2}{9} = 1 \quad \dots (2)$ Solving simultaneously, we isolate coordinates: (x,y) equiv left( pm fracsqrt6sqrt5, pm frac6sqrt5 right)$(x,y) \equiv \left( \pm \frac{\sqrt{6}}{\sqrt{5}}, \pm \frac{6}{\sqrt{5}} \right)$ ### Step 3: Calculate Area of Quadrilateral The four symmetrical intersection points form a rectangle of dimension 2x times 2y$2x \times 2y$: textArea = 2left(fracsqrt6sqrt5right) times 2left(frac6sqrt5right) = frac24sqrt65$\text{Area} = 2\left(\frac{\sqrt{6}}{\sqrt{5}}\right) \times 2\left(\frac{6}{\sqrt{5}}\right) = \frac{24\sqrt{6}}{5}$ ### Pattern Recognition When two ellipses centered at origin intersect symmetrically across the axes, the intersection area is always a rectangle of area 4|x cdot y|$4|x \cdot y|$ computed directly from roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula - Eccentricity of hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: e_h = sqrt1 + fracb^2a^2$e_h = \sqrt{1 + \frac{b^2}{a^2}}$ - Eccentricity of ellipse fracx^2B^2 + fracy^2A^2 = 1$\frac{x^2}{B^2} + \frac{y^2}{A^2} = 1$ (where A > B$A > B$): e_e = sqrt1 - fracB^2A^2$e_e = \sqrt{1 - \frac{B^2}{A^2}}$ ### Core Logic Let's rewrite both equations in their standard forms: 1. **Hyperbola:** x^2 - y^2csc^2theta = 5 implies fracx^25 - fracy^25sin^2theta = 1$x^2 - y^2\csc^2\theta = 5 \implies \frac{x^2}{5} - \frac{y^2}{5\sin^2\theta} = 1$ Here, a^2 = 5$a^2 = 5$ and b^2 = 5sin^2theta$b^2 = 5\sin^2\theta$. e_h = sqrt1 + frac5sin^2theta5 = sqrt1 + sin^2theta$e_h = \sqrt{1 + \frac{5\sin^2\theta}{5}} = \sqrt{1 + \sin^2\theta}$ 2. **Ellipse:** x^2csc^2theta + y^2 = 5 implies fracx^25sin^2theta + fracy^25 = 1$x^2\csc^2\theta + y^2 = 5 \implies \frac{x^2}{5\sin^2\theta} + \frac{y^2}{5} = 1$ Since 0 < theta < pi/2$0 < \theta < \pi/2$, we know 0 < sintheta < 1 implies 5sin^2theta < 5$0 < \sin\theta < 1 \implies 5\sin^2\theta < 5$. Thus, the major axis is along the y-axis, meaning A^2 = 5$A^2 = 5$ and B^2 = 5sin^2theta$B^2 = 5\sin^2\theta$. e_e = sqrt1 - frac5sin^2theta5 = sqrt1 - sin^2theta = costheta$e_e = \sqrt{1 - \frac{5\sin^2\theta}{5}} = \sqrt{1 - \sin^2\theta} = \cos\theta$ ### Step 1: Setting up the Equation and Solving Given e_h = sqrt7e_e$e_h = \sqrt{7}e_e$, squaring both sides yields: e_h^2 = 7e_e^2$e_h^2 = 7e_e^2$ 1 + sin^2theta = 7(1 - sin^2theta)$1 + \sin^2\theta = 7(1 - \sin^2\theta)$ 1 + sin^2theta = 7 - 7sin^2theta$1 + \sin^2\theta = 7 - 7\sin^2\theta$ 8sin^2theta = 6 implies sin^2theta = frac34$8\sin^2\theta = 6 \implies \sin^2\theta = \frac{3}{4}$ Since theta in (0, pi/2)$\theta \in (0, \pi/2)$: sintheta = fracsqrt32 implies theta = fracpi3$\sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$ ### Pattern Recognition Sees: Conic parameters tied dynamically to trigonometry. Trap: In the ellipse equation, do not automatically assume the x-axis holds the major axis. Since sin^2theta < 1$\sin^2\theta < 1$, the denominator under y^2$y^2$ is larger, making it a vertical ellipse. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Trigonometric Functions

### Related Formula - Eccentricity of an ellipse (a>b$a>b$): e_E^2 = 1 - fracb^2a^2$e_E^2 = 1 - \frac{b^2}{a^2}$ - Eccentricity of a hyperbola: e_H^2 = 1 + fracb^2a^2$e_H^2 = 1 + \frac{b^2}{a^2}$ ### Core Logic Given the eccentricity of the ellipse is e_E = frac1sqrt2$e_E = \frac{1}{\sqrt{2}}$: e_E^2 = 1 - fracb^2a^2 implies left(frac1sqrt2right)^2 = 1 - fracb^2a^2$e_E^2 = 1 - \frac{b^2}{a^2} \implies \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{b^2}{a^2}$ frac12 = 1 - fracb^2a^2 implies fracb^2a^2 = frac12$\frac{1}{2} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{1}{2}$ ### Step 1: Finding Hyperbola Eccentricity The required equation asks for the square of the eccentricity of the conjugate parameter hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: e_H^2 = 1 + fracb^2a^2$e_H^2 = 1 + \frac{b^2}{a^2}$ Substituting the derived ratio fracb^2a^2 = frac12$\frac{b^2}{a^2} = \frac{1}{2}$: e_H^2 = 1 + frac12 = frac32$e_H^2 = 1 + \frac{1}{2} = \frac{3}{2}$ Thus, the square of the eccentricity is frac32$\frac{3}{2}$. ### Pattern Recognition Sees: Linked parameters between ellipse and hyperbola. Shortcut: Notice that information about the latus rectum length (sqrt14$\sqrt{14}$) is a deliberate distractor! Since the definition of hyperbola eccentricity squared (e_H^2 = 1 + b^2/a^2$e_H^2 = 1 + b^2/a^2$) relies solely on the dimensionless ratio fracb^2a^2$\frac{b^2}{a^2}$, you don't need to compute absolute scales for a$a$ or b$b$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections