Let H₁: x²a²- y²b²=1$H_{1}:\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and H₂:- x²A²+ y²B²=1$H_{2}:-\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$ be two hyperbolas having length of latus rectums 15√(2)$15\sqrt{2}$ and 12√(5)$12\sqrt{5}$ respectively. Let their eccentricities be e₁=√((5)/(2))$e_{1}=\sqrt{\frac{5}{2}}$ and e₂$e_{2}$ respectively. If the product of the lengths of their transverse axes is 100√(10)$100\sqrt{10}$ then 25e₂²$25e_{2}^{2}$ is equal to \_\_\_\_.

Related Formula

For a circle in the third quadrant touching both coordinate axes, the center layout is (-r, -r)$(-r, -r)$ and equation looks like:

For external contact between circles C₁$C_1$ and C₂$C_2$, the distance between centers equals the sum of their radii:

Core Logic

Circle C₁$C_1$ has radius r₁ = 3$r_1 = 3$ and touches both axes in the third quadrant, so its center is A(-3, -3)$A(-3, -3)$. Circle C₂$C_2$ has center B(1, 3)$B(1, 3)$.

The distance between centers A$A$ and B$B$ is:

Step 1: Determine Radius of Circle 2

Step 2: Locate the Contact Point via Section Formula

The point of contact P(α, β)$P(\alpha, \beta)$ divides the line segment joining centers A(-3, -3)$A(-3, -3)$ and B(1, 3)$B(1, 3)$ internally in the ratio r₁ : r₂ = 3 : (2√(13) - 3)$r_1 : r_2 = 3 : (2\sqrt{13} - 3)$.

Using the internal section formula:

Step 3: Calculate the Difference Value

Find (β - α)²$(\beta - \alpha)^2$:

Comparing with (m)/(n)$\frac{m}{n}$ where (m, n) = 1$\gcd(m, n) = 1$ gives m = 9, n = 13$m = 9, n = 13$.

Pattern Recognition

Notice that computing (β - α)$(\beta - \alpha)$ directly cancels out the irrational √(13)$\sqrt{13}$ term from the numerator before squaring, saving a significant amount of tedious arithmetic expansion.

Chapter Mix

Class 11 Mathematics: Coordinate Geometry Class 11 Mathematics: Circles

Related Formula

For a standard hyperbola:

• Focus positions are (± ae, 0)$(\pm ae, 0)$.
• Length of semi-latus rectum is (b²)/(a)$\frac{b^2}{a}$.
• Eccentricity identity linkage: b² = a²(e² - 1) a²e² = a² + b²$b^2 = a^2(e^2 - 1) \implies a^2e^2 = a^2 + b^2$.

Core Logic

Given focus F₁ ≡ (-ae, 0) ≡ P(-3, 0)$F_1 \equiv (-ae, 0) \equiv P(-3, 0)$, so ae = 3$ae = 3$. The other focus is F₂ ≡ (ae, 0) ≡ (3, 0)$F_2 \equiv (ae, 0) \equiv (3, 0)$.

The latus rectum passes vertically through F₂$F_2$, with endpoints L₁(ae, (b²)/(a))$L_1\left(ae, \frac{b^2}{a}\right)$ and L₂(ae, -(b²)/(a))$L_2\left(ae, -\frac{b^2}{a}\right)$. This segment subtends a right angle at P(-ae, 0)$P(-ae, 0)$. By symmetry, the top half angle at P$P$ must be exactly 45^°$45^\circ$.

Step 1: Set Up Slope Relationship

Step 2: Solve the Quadratic Excentricity Equation

Substitute ae = 3$ae = 3$ and b² = 6a$b^2 = 6a$ into the eccentricity identity a²e² = a² + b²$a^2e^2 = a^2 + b^2$:

Solving for a$a$ using the quadratic formula (taking the positive root since a > 0$a > 0$):

Step 3: Evaluate product and sum coefficients

Now compute a²b²$a^2b^2$:

Matching with α√(2) - β$\alpha\sqrt{2} - \beta$ gives:

Calculate the final required sum:

Pattern Recognition

Recognizing that the right angle subtended at the opposite focus implies a perfect (45^°)$\tan(45^\circ)$ right triangle instantly yields the key linear constraint b² = 2a(ae)$b^2 = 2a(ae)$, avoiding the need for lengthy distance-formula tracking.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

Core Logic

Extract the coordinate center of the target circle, substitute it directly to locate the missing parameter p$p$, and resolve eccentricity metrics.

Step 1: Extract Circle Metric Values

For circle x² + y² - 2x - 4y - 11 = 0$x^{2} + y^{2} - 2x - 4y - 11 = 0$:

Step 2: Standardize Ellipse Formulation

Ellipse passes through point C(1,2)$C(1,2)$:

Standard model form: (x²)/(4/3) + (y²)/(16) = 1$\frac{x^2}{4/3} + \frac{y^2}{16} = 1$ (b > a$b > a$, vertical configuration axis).

Step 3: Evaluate Product Chain

Focal distance elements at ordinate coordinate height k=2$k=2$ are bounded by b ± ek$b \pm ek$:

Target evaluation expression response string:

Pattern Recognition

Pay attention to whether b > a$b > a$ or a > b$a > b$ when analyzing ellipse forms. Focal distance definitions swap directions immediately across major horizontal/vertical configurations.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Related Formula

Core Logic

Establish the tangent vector expression at the parabola intersection mark. Since this path line functions as a shared contact tangent boundaries sheet for the circular arc, impose radius equations.

Step 1: Derive Shared Parabola Tangent Line

Tangent line profile for y² = 9x$y^2 = 9x$ at coordinate indicator (4,6)$(4,6)$:

Step 2: Build Geometric Metric Connections

Circle touches axis at (a,0)$(a,0)$, mapping coordinates center directly to C(a,r)$C(a,r)$. Perpendicular boundary constraint steps require:

Step 3: Solve for Radius Matrix Bounds

Enforce circle equation intersection constraint profile (x-a)² + (y-r)² = r²$(x-a)^2 + (y-r)^2 = r^2$ at point (4,6)$(4,6)$:

Evaluating the target systems from structural logic tracks rejects positive value parameters, providing:

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Pattern Recognition

Shared tangent elements connect independent conic fields. Locating circular center boundaries using axial coordinate tracking simplifies secondary equations.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Related Formula

Equation of a chord of a conic section with a given midpoint (x₁, y₁)$(x_1, y_1)$ is:

T = S₁$T = S_1$

Core Logic

Given midpoint M((5)/(2), (1)/(2))$M\left(\frac{5}{2}, \frac{1}{2}\right)$ and ellipse (x²)/(9) + (y²)/(4) = 1$\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Write T$T$ and S₁$S_1$ terms:

Equating both sides:

Step 1: Simplify to Standard Form

Multiply the entire equation by 144 to eliminate fractions:

40x + 18y = 109$40x + 18y = 109$

Comparing this directly with α x + β y = 109$\alpha x + \beta y = 109$ provides:

Pattern Recognition

Whenever you see 'chord whose midpoint is given', write T = S₁$T = S_1$ automatically. Match coefficients directly at the final step after equating constant integers.

Chapter Mix

Class 11 Mathematics: Conic Sections