Let the line L -:√(2)x+y=α$-:\sqrt{2}x+y=\alpha$ pass through the point of the intersection P (in the first quadrant) of the circle x²+y²=3$x^{2}+y^{2}=3$ and the parabola x²=2y$x^{2}=2y$. Let the line L touch two circles C₁$C_{1}$ and C₂$C_{2}$ of equal radius 2√(3)$2\sqrt{3}$. If the centres Q₁ and Q₂$Q_{2}$ of the circles C₁$C_{1}$ and C₂$C_{2}$ lie on the y-axis, then the square of the area of the triangle PQ₁Q₂$PQ_{1}Q_{2}$ is equal to

Related Formula

Latus rectum length = (2b²)/(a) = 8 b² = 4a$= \frac{2b^2}{a} = 8 \implies b^2 = 4a$. Focal length = 2ae = 2√(a² + b²)$= 2ae = 2\sqrt{a^2 + b^2}$.

Core Logic

Substitute P(10, 2√(15))$P(10, 2\sqrt{15})$ and b² = 4a$b^2 = 4a$ into hyperbola equation:

Thus, b² = 20 b = √(20)$b^2 = 20 \implies b = \sqrt{20}$.

Step 1: Calculate Focal Distance and Area

Focal distance SS' = 2ae = 2 √(a² + b²) = 2 √(25 + 20) = 6√(5)$SS' = 2ae = 2 \sqrt{a^2 + b^2} = 2 \sqrt{25 + 20} = 6\sqrt{5}$. Area of Δ PSS' = (1)/(2) × base × height = (1)/(2) (6√(5)) (2√(15)) = 30√(3) = A$\Delta PSS' = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (6\sqrt{5}) (2\sqrt{15}) = 30\sqrt{3} = A$.

Step 2: Square of Area

Pattern Recognition

Use latus rectum relation to reduce hyperbola parameter to single variable quadratic.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

For ellipse (x²)/(a²) + (y²)/(b²) = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: SP + S'P = 2a = 10$SP + S'P = 2a = 10$, e = √(1 - b²/a²) = √(1 - 9/25) = 4/5$e = \sqrt{1 - b^2/a^2} = \sqrt{1 - 9/25} = 4/5$. Focal distances: SP = a - eα = 5 - (4)/(5)α$SP = a - e\alpha = 5 - \frac{4}{5}\alpha$, S'P = a + eα = 5 + (4)/(5)α$S'P = a + e\alpha = 5 + \frac{4}{5}\alpha$.

Core Logic

Using algebraic identity:

Substitute focal distance formula:

Step 1: Calculate beta^2 and Sum

Substitute α²$\alpha^2$ into ellipse equation (α²)/(25) + (β²)/(9) = 1$\frac{\alpha^2}{25} + \frac{\beta^2}{9} = 1$:

Pattern Recognition

Express (SP)² + (S'P)² - SP · S'P$(SP)^2 + (S'P)^2 - SP \cdot S'P$ in terms of (SP+S'P)$(SP+S'P)$ to determine SP · S'P$SP \cdot S'P$ instantly.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Section formula for internal division in ratio m:n$m:n$:

Core Logic

Let chord endpoint be Q(t², 2t)$Q(t^2, 2t)$. Midpoint M(h,k)$M(h,k)$ of OQ$OQ$:

So curve S$S$ is y² = 2x$y^2 = 2x$.

Now P$P$ lies on S: y² = 2x$S: y^2 = 2x$, so P = ((t²)/(2), t)$P = \left(\frac{t^2}{2}, t\right)$. Point R(h,k)$R(h,k)$ divides OP$OP$ in ratio 3:1$3:1$:

Step 1: Section Formula Application

From k = (3t)/(4) t = (4k)/(3)$k = \frac{3t}{4} \implies t = \frac{4k}{3}$. Substitute into h$h$:

Pattern Recognition

Parametrize midpoint curve S$S$, then re-apply section ratio to derive final locus equation.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Core Logic

For the domain of the given logarithmic function, the argument of the innermost logarithm must be strictly greater than 1 because of the nested logs:

Thus, 4 < x < 5$4 < x < 5$. Therefore, the domain interval is (4, 5)$(4, 5)$, yielding m = 4$m = 4$ and n = 5$n = 5$.

Step 1: Hyperbola Properties

Given eccentricity e = (n)/(3) = (5)/(3)$e = \frac{n}{3} = \frac{5}{3}$:

Given the length of the latus rectum is (8m)/(3)$\frac{8m}{3}$:

Since (b)/(a) = (4)/(3)$\frac{b}{a} = \frac{4}{3}$, we get a = 3$a = 3$.

Step 2: Final Calculation

We need to find b² - a²$b^2 - a^2$:

Pattern Recognition

Nested logarithmic domains require unpacking from the outside in: ₐ(X) > 0 ⇒ X > 1$\log_a(X) > 0 \Rightarrow X > 1$. Linking function domains to coordinate geometry parameters is a standard JEE cross-topic pattern.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Relations and Functions

Related Formula

Core Logic

Find the intersection points of the line y = x + 1$y = x + 1$ and the ellipse (x²)/(2) + y² = 1$\frac{x^2}{2} + y^2 = 1$.

This gives x = 0$x = 0$ or x = -(4)/(3)$x = -\frac{4}{3}$.

Step 1: Calculate Intersection Points

For x = 0$x = 0$, y = 1 ⇒ A(0, 1)$y = 1 \Rightarrow A(0, 1)$. For x = -(4)/(3)$x = -\frac{4}{3}$, y = -(4)/(3) + 1 = -(1)/(3) ⇒ B(-(4)/(3), -(1)/(3))$y = -\frac{4}{3} + 1 = -\frac{1}{3} \Rightarrow B\left(-\frac{4}{3}, -\frac{1}{3}\right)$.

Step 2: Find Angle at the Origin

Let O(0,0)$O(0,0)$ be the center of the ellipse. The angle made by segment AB$AB$ at O$O$ is ∠ AOB$\angle AOB$. The slope of OA$OA$ is m₁ = (1 - 0)/(0 - 0) = ∞$m_1 = \frac{1 - 0}{0 - 0} = \infty$ (which means OA$OA$ is along the y-axis, angle is π/2$\pi/2$). The slope of OB$OB$ is m₂ = (-1/3 - 0)/(-4/3 - 0) = (1)/(4)$m_2 = \frac{-1/3 - 0}{-4/3 - 0} = \frac{1}{4}$. The angle of OB$OB$ with the positive x-axis is θ = ⁻¹((1)/(4))$\theta = \tan^{-1}\left(\frac{1}{4}\right)$. The total angle ∠ AOB$\angle AOB$ is (π)/(2) + θ = (π)/(2) + ⁻¹((1)/(4))$\frac{\pi}{2} + \theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$.

Pattern Recognition

When solving line-conic intersection, explicit extraction of points (A, B)$(A, B)$ is often simpler than using homogenization if the intersection yields simple rational or integer coordinates.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines