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

The condition for a line y = mx + p$y = mx + p$ to be tangent to an ellipse (x²)/(a²) + (y²)/(b²) = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

The coordinate of the point of contact is given by (-(a²m)/(p), (b²)/(p))$\left(-\frac{a^2m}{p}, \frac{b^2}{p}\right)$.

Core Logic

Given the ellipse parameter values a² = 16$a^2 = 16$ and b² = 9$b^2 = 9$, and tangent line slope m = 1$m = 1$:

Thus, the two values of p$p$ are 5$5$ and -5$-5$. The points of contact A$A$ and B$B$ are:

• For p = 5$p = 5$: A = (-(16(1))/(5), (9)/(5)) = (-(16)/(5), (9)/(5))$A = \left(-\frac{16(1)}{5}, \frac{9}{5}\right) = \left(-\frac{16}{5}, \frac{9}{5}\right)$
• For p = -5$p = -5$: B = (-(16(1))/(-5), (9)/(-5)) = ((16)/(5), -(9)/(5))$B = \left(-\frac{16(1)}{-5}, \frac{9}{-5}\right) = \left(\frac{16}{5}, -\frac{9}{5}\right)$

Step 1: Intersecting line with Ellipse

The line y = x$y = x$ intersects the ellipse (x²)/(16) + (y²)/(9) = 1$\frac{x^2}{16} + \frac{y^2}{9} = 1$:

Since y = x$y = x$, the intersection points C$C$ and D$D$ are:

Step 2: Calculating Quadrilateral Area

The area of quadrilateral ABCD$ABCD$ with vertices mapped symmetrically can be computed using the standard coordinate determinant matrix layout formula:

Pattern Recognition

Notice that the tangent lines are parallel and symmetric (p = ± 5$p = \pm 5$), and the intersecting line passes through the origin. This symmetry creates a geometric parallelogram, simplifying your area calculation by doubling the area of triangle ABD$ABD$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

Distance between foci = 2be$= 2be$ (for vertical major axis). Length of Latus Rectum (L.R.) = (2a²)/(b)$(\text{L.R.}) = \frac{2a^2}{b}$.

Core Logic

Foci are F₁(2,5)$F_1(2,5)$ and F₂(2,-3)$F_2(2,-3)$. The x$x$-coordinates are identical, indicating a vertical ellipse. Distance between foci:

Given eccentricity e = (4)/(5)$e = \frac{4}{5}$:

Step 1: Calculate Minor Axis length

Use eccentricity relation:

Step 2: Evaluate Latus Rectum

Pattern Recognition

Always verify axis orientation (horizontal vs vertical) from coordinates before blindly substituting into standard formulas. Symmetrical components match axis lengths parameters directly.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

Parametric equation of a straight line:

Core Logic

Step 1: Find point P(x₁,y₁)$P(x_1,y_1)$ on circle C$C$. Equation of tangent at P$P$ on x² + y² - 2y - 1 = 0$x^2 + y^2 - 2y - 1 = 0$ is xx₁ + y(y₁ - 1) - (y₁ + 1) = 0$xx_1 + y(y_1 - 1) - (y_1 + 1) = 0$. Comparing with line x + y = 3 (x₁)/(1) = (y₁ - 1)/(1) = (y₁ + 1)/(3)$x + y = 3 \implies \frac{x_1}{1} = \frac{y_1 - 1}{1} = \frac{y_1 + 1}{3}$. Solving gives x₁ = 1, y₁ = 2$x_1 = 1, y_1 = 2$. Thus, P = (1, 2)$P = (1, 2)$.

Step 1: Use Line Parametrics for Q and R

Line x + y = 3$x + y = 3$ makes an angle θ = 135°$\theta = 135^{\circ}$ with the positive x-axis. Using parametric distances from P(1,2)$P(1,2)$ with r = PQ = 2√(2)3$r = PQ = \frac{2\sqrt{2}}{3}$:

Substitute r = 2√(2)3$r = \frac{2\sqrt{2}}{3}$: For Q$Q$: x₂ = 1 + (2)/(3) = (5)/(3), y₂ = 2 - (2)/(3) = (4)/(3)$x_2 = 1 + \frac{2}{3} = \frac{5}{3}, y_2 = 2 - \frac{2}{3} = \frac{4}{3}$. For R$R$: x₃ = 1 - (2)/(3) = (1)/(3), y₃ = 2 + (2)/(3) = (8)/(3)$x_3 = 1 - \frac{2}{3} = \frac{1}{3}, y_3 = 2 + \frac{2}{3} = \frac{8}{3}$.

Step 2: Evaluate Final Expression

Calculate the products: x₁y₁ = 1 × 2 = 2$x_1y_1 = 1 \times 2 = 2$ x₂y₂ = (5)/(3) × (4)/(3) = (20)/(9)$x_2y_2 = \frac{5}{3} \times \frac{4}{3} = \frac{20}{9}$ x₃y₃ = (1)/(3) × (8)/(3) = (8)/(9)$x_3y_3 = \frac{1}{3} \times \frac{8}{3} = \frac{8}{9}$

Pattern Recognition

Parametric distance equations are perfect for lines containing midpoints. This approach bypasses calculating the individual ellipse equations a², b²$a^2, b^2$ completely.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Conic Sections

Related Formula

Length of latus rectum of an ellipse and its eccentricity relation are:

Core Logic

Given length of LR = 10 (2b²)/(a) = 10 b² = 5a (i)$\text{LR} = 10 \implies \frac{2b^2}{a} = 10 \implies b^2 = 5a \quad \dots \text{(i)}$

Now, let's find the minimum value of f(t) = t² + t + (11)/(12)$f(t) = t^2 + t + \frac{11}{12}$. Differentiating: f'(t) = 2t + 1 = 0 t = -(1)/(2)$f'(t) = 2t + 1 = 0 \implies t = -\frac{1}{2}$.

Step 1: Solve for a and b

Using eccentricity formula:

Equating (i) and (ii):

Then from (i):

Hence, a² = 81$a^2 = 81$.

Step 2: Calculate a^2 + b^2

Pattern Recognition

A quadratic function at²+bt+c$at^2+bt+c$ reaches its extreme value at t = -(b)/(2a)$t = -\frac{b}{2a}$. Using this layout avoids full calculus derivation steps.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

Related Formula

Eccentricity for ellipse (a$a) and hyperbola are calculated as:

Core Logic

Since b < 5$b < 5$, for the ellipse (x²)/(b²) + (y²)/(25) = 1$\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the major axis is along the y$y$-axis:

For the hyperbola (x²)/(16) - (y²)/(b²) = 1$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$:

Given e₁² e₂² = 1$e_1^2 e_2^2 = 1$:

Step 1: Calculate Foci Locations

Substituting b² = 9$b^2 = 9$: Ellipse foci: ae₁ = 5 · √(1 - (9)/(25)) = 5 · (4)/(5) = 4$ae_1 = 5 \cdot \sqrt{1 - \frac{9}{25}} = 5 \cdot \frac{4}{5} = 4$. Foci lie along y$y$-axis: (0, ± 4)$(0, \pm 4)$. Hyperbola foci: ae₂ = 4 · √(1 + (9)/(16)) = 4 · (5)/(4) = 5$ae_2 = 4 \cdot \sqrt{1 + \frac{9}{16}} = 4 \cdot \frac{5}{4} = 5$. Foci lie along x$x$-axis: (± 5, 0)$(\pm 5, 0)$.

Step 2: Construct the New Ellipse

The new ellipse passes through (0, ± 4)$(0, \pm 4)$ and (± 5, 0)$(\pm 5, 0)$. Thus, its semi-major axis is A = 5$A = 5$ along the x$x$-axis and semi-minor axis is B = 4$B = 4$ along the y$y$-axis:

Pattern Recognition

When a conic passes through points directly on the axes like (± alpha, 0)$(\pm alpha, 0)$ and (0, ± β)$(0, \pm \beta)$, those points immediately represent the semi-axes values A$A$ and B$B$.

Chapter Mix

Class 11 Mathematics: Conic Sections