Let E₁: (x²)/(9) + (y²)/(4) = 1$E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$ be an ellipse. Ellipses Eᵢ$E_i$ 's are constructed such that their centres and eccentricities are same as that of E₁$E_1$ , and the length of minor axis of Eᵢ$E_i$ is the length of major axis of Eᵢ₊₁$E_{i+1}$ ( i ≥ 1$i \ge 1$ ). If Aᵢ$A_i$ is the area of the ellipse Eᵢ$E_i$ , then (5)/(pi) ( Σi=1∞ Aᵢ )$\frac{5}{pi} \left( \sum_{i=1}^{\infty} A_i \right)$ , is equal to ....

Related Formula

Core Logic

From the given coordinates on the major axis (y = -2$y = -2$): Center C(1, -2)$C(1, -2)$, Focus F₁(3, -2)$F_1(3, -2)$, Vertex A₁(5, -2)$A_1(5, -2)$. Distance from center to vertex, CA₁ = a = 5 - 1 = 4$CA_1 = a = 5 - 1 = 4$. Distance from center to focus, CF₁ = ae = 3 - 1 = 2$CF_1 = ae = 3 - 1 = 2$.

Execution

Calculate eccentricity e$e$:

Use alternate formula for Latus Rectum:

Pattern Recognition

Aligning focus, center, and vertex along a constant y-axis implies a standard shifted ellipse where absolute differences in x-coordinates yield standard parameters (a$a$ and ae$ae$) directly.

Chapter Mix

Class 11 Maths: Conic Sections

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Core Logic

For Ellipse E: (x²)/(144) + (y²)/(169) = 1$E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ a² = 144, b² = 169$a^2 = 144, b^2 = 169$. Since b > a$b > a$, the major axis is along the y-axis. Eccentricity e' = √(1 - (144)/(169)) = √((25)/(169)) = (5)/(13)$e' = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$. Foci of ellipse = (0, ± be') = (0, ± 13 × (5)/(13)) = (0, ± 5)$(0, \pm be') = (0, \pm 13 \times \frac{5}{13}) = (0, \pm 5)$.

Execution

For Hyperbola H: (y²)/(λ²) - (x²)/(16) = 1$H: \frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$ Foci of conjugate hyperbola are (0, ± λ e)$(0, \pm \lambda e)$. Equating foci: λ e = 5$\lambda e = 5$.

Eccentricity of hyperbola, e = (5)/(3)$e = \frac{5}{3}$. Length of latus rectum of hyperbola, L = (2(16))/(λ) = (32)/(3)$L = \frac{2(16)}{\lambda} = \frac{32}{3}$.

Calculate 24(e + L)$24(e + L)$:

Pattern Recognition

Confocal conics usually align along the same major axis. Notice the -1$-1$ on the RHS of the hyperbola equation indicates a conjugate hyperbola orienting it vertically to match the b>a$b>a$ ellipse.

Chapter Mix

Class 11 Maths: Conic Sections

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Core Logic

We set up an algebraic equation relating b$b$, a$a$, and e$e$ from the given geometric condition, then substitute it into the eccentricity identity.

Step 1: Set up the geometric relation

Given that 2b = (1)/(4) (2ae)$2b = \frac{1}{4} (2ae)$:

Square both sides:

Step 2: Solve for eccentricity

Using the eccentricity relation:

Pattern Recognition

Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing b/a$b/a$ as a function of e$e$ allows direct solving of the eccentricity.

Chapter Mix

Class 11 Mathematics: Conic Sections

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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 y² = 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 t₁², 2a t₁) = (1, -4)$(a t_1^2, 2a t_1) = (1, -4)$:

Since PQ$PQ$ is a focal chord, the parametric points are coupled:

Now, calculate the coordinates of Q$Q$:

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 λ : 1$\lambda : 1$. Using the y$y$-coordinate of the section formula:

Thus, the focus S$S$ divides the chord internally in the ratio 1:4$1:4$. Since (1, 4) = 1$\gcd(1, 4) = 1$, we have m = 1$m = 1$ and n = 4$n = 4$:

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: (1)/(SP) + (1)/(SQ) = (1)/(a)$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For a standard hyperbola: Focus: (± ae, 0)$(\pm ae, 0)$ Directrix: x = ± (a)/(e)$x = \pm \frac{a}{e}$ Eccentricity relation: (ae)² = a² + b²$(ae)^2 = a^2 + b^2$ Length of latus rectum: l = (2b²)/(a)$l = \frac{2b^2}{a}$

Core Logic

Given ae = √(10)$ae = \sqrt{10}$ and (a)/(e) = 9√(10)$\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these gives a²$a^2$, which determines both parameters.

Step 1: Find a and e

Substitute a = 3$a = 3$ into ae = √(10)$ae = \sqrt{10}$:

Step 2: Find b and l

Using (ae)² = a² + b²$(ae)^2 = a^2 + b^2$:

Then the length of latus rectum l$l$ is:

Step 3: Evaluate Final Expression

Calculate 9(e² + l)$9(e^2 + l)$:

Pattern Recognition

Multiplying focus location by directrix location immediately eliminates e$e$, giving a²$a^2$ directly. Once a²$a^2$ is known, b²$b^2$ follow seamlessly via (ae)² = a²+b²$(ae)^2 = a^2+b^2$.

Chapter Mix

Class 11 Mathematics: Conic Sections