A line passing through the point P(√(5), √(5))$P(\sqrt{5}, \sqrt{5})$ intersects the ellipse (x²)/(36) + (y²)/(25) = 1$\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A$A$ and B$B$ [cite: 567] such that (PA) · (PB)$(PA) \cdot (PB)$ is maximum. Then 5(PA² + PB²)$5(PA^{2} + PB^{2})$ is equal to

Related Formula

Length of a chord with perpendicular distance d$d$ from the center of a circle of radius r$r$ is:

Core Logic

Points A(4,2)$A(4,2)$ and B(0,2)$B(0,2)$ have the same y-coordinate, meaning chord AB$AB$ is horizontal. The perpendicular bisector of a horizontal chord is vertical.

Midpoint of AB$AB$ is M(2,2)$M(2,2)$. Thus, the vertical line passing through the center is x = 2$x = 2$.

Step 1: Identify Center and Radius

Since the center lies on the line 3x + 2y + 2 = 0$3x + 2y + 2 = 0$, substitute x = 2$x = 2$ to find the y-coordinate:

So, Center O = (2, -4)$O = (2, -4)$.

Calculate radius r$r$ using point B(0,2)$B(0,2)$:

Step 2: Find Target Chord Length

The targeted chord has a given midpoint N(1,2)$N(1,2)$. Distance from center O(2,-4)$O(2,-4)$ to N(1,2)$N(1,2)$:

Pattern Recognition

Points sharing a coordinate define standard vertical or horizontal perpendicular configurations immediately. Always exploit geometrical configurations before jumping into standard circle equations.

Chapter Mix

Class 11 Mathematics: Circles

Related Formula

Properties of focal chord parameter metrics in parabolas y² = 4ax$y^2 = 4ax$:

Distance to the directrix property:

Core Logic

Given parabola y² = 12x a = 3$y^2 = 12x \implies a = 3$. Focus S = (3, 0)$S = (3, 0)$. Set up focal segments product equation:

Solving for t²$t^2$:

Step 1: Compute Endpoint Coordinate Bounds

Choosing t = - √(3)2$t = -\frac{\sqrt{3}}{2}$ allows defining both chord coordinates symmetrically:

Step 2: Derive Circle Equation

Write the diameter circle form equation:

Multiply by 64 to clear the fractions and match the given equation template structure:

Comparing directly with 64x² + 64y² - α x - 64√(3)y = β$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$ yields:

Pattern Recognition

The distance from focal chord endpoints to the focus equals their perpendicular distance to the directrix. This property connects parameter metrics to geometric lengths cleanly.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Related Formula

Area of a trapezium is given by:

Core Logic

Let the coordinates of the vertices be parameterized on the parabola y² = 4x$y^2 = 4x$. Since AD$AD$ and BC$BC$ are parallel to the y-axis, the coordinates take the form: A(at₁², 2at₁)$A(at_1^2, 2at_1)$ and D(at₁², -2at₁)$D(at_1^2, -2at_1)$ B(at₂², 2at₂)$B(at_2^2, 2at_2)$ and C(at₂², -2at₂)$C(at_2^2, -2at_2)$

Given a=1$a=1$, the points simplify accordingly.

Step 1: Using Diagonal Properties

The length of diagonal AC$AC$ passing through focal point (1,0)$(1,0)$ implies focal chord properties:

Step 2: Finding Coordinates and Area

Substituting t₁ = 2$t_1 = 2$, we get: A((1)/(2), 1), D((1)/(4), -1), B(4, 4), C(4, -4)$A\left(\frac{1}{2}, 1\right), D\left(\frac{1}{4}, -1\right), B(4, 4), C(4, -4)$

Evaluating the area formula:

Pattern Recognition

Focal chords of parabolas always satisfy t₁ t₂ = -1$t_1 t_2 = -1$. Recognizing the passage through (1,0)$(1,0)$ unlocks quick parametric simplifications.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Circle intercepts standard form templates:

Core Logic

Since the circle touches the x-axis at (a,0)$(a,0)$ and lies entirely below it, its center is located at (a, -p)$(a, -p)$ where p$p$ matches its radius r$r$.

By Pythagoras' theorem:

Step 1: Translating to General Equation Form

The explicit standard equation is (x-a)² + (y+p)² = r²$(x-a)^2 + (y+p)^2 = r^2$. Expanding it out:

Comparing this directly to x² + y² - α x + β y + γ = 0$x^2 + y^2 - \alpha x + \beta y + \gamma = 0$ yields: α = 2a$\alpha = 2a$, β = 2p$\beta = 2p$, and γ = a²$\gamma = a^2$.

Step 2: Evaluating the Target Mapped Ordered Pair

Isolating b²$b^2$ using the parametric radius dimensions:

Thus, the mapped ordered pair (2a, b²)$(2a, b^2)$ evaluates directly to (α, β² - 4γ)$(\alpha, \beta^2 - 4\gamma)$.

Pattern Recognition

Tangency conditions fix center parameters to match radius scale sizes instantly, reducing variable overhead in coordinate transformations.

Chapter Mix

Class 11 Maths: Circles

Related Formula

Area of an ellipse with semi-axes a$a$ and b$b$:

Core Logic

Calculate the constant eccentricity e$e$ from the initial ellipse E₁$E_1$:

For any subsequent ellipse E₂$E_2$, its major axis equals the minor axis of E₁$E_1$ (2b₁ = 4 a₂ = 2$2b_1 = 4 \implies a_2 = 2$). Since eccentricity remains constant:

Step 1: Finding the Area Sequence Terms

Evaluate the area values for the initial ellipses: A₁ = π · 3 · 2 = 6π$A_1 = \pi \cdot 3 \cdot 2 = 6\pi$ A₂ = π · 2 · (4)/(3) = (8π)/(3)$A_2 = \pi \cdot 2 \cdot \frac{4}{3} = \frac{8\pi}{3}$

The areas form an infinite geometric progression with a common ratio r = (4)/(9)$r = \frac{4}{9}$.

Step 2: Summing the Infinite Geometric Series

Evaluating the final scaling formula:

Pattern Recognition

Iterative dimensional scaling creates geometric progressions where the ratio equals the square of the linear scaling factor.

Chapter Mix

Class 11 Maths: Conic Sections