Let y² = 12x$y^2 = 12x$ the parabola and S$S$ be its focus. Let PQ$PQ$ be a focal chord of the parabola such that (SP) (SQ) = (147)/(4)$(SQ) = \frac{147}{4}$. Let C$C$ be the circle described taking PQ$PQ$ as a diameter. If the equation of a circle C$C$ is 64x² + 64y² - α x - 64√(3)y = β$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$, then \beta - \alpha is equal to

Related Formula

Core Logic

Given line: α x + 2y = 1 y = (1 - α x)/(2)$\alpha x + 2y = 1 \implies y = \frac{1 - \alpha x}{2}$. Given hyperbola: x² - 9y² = 9$x^2 - 9y^2 = 9$.

Substitute the expression for y$y$ into the hyperbola's equation:

Step 1: Solving for Discriminant

Multiply by 4:

For the line to NOT intersect the hyperbola, the quadratic must yield non-real roots, meaning Discriminant D < 0$D < 0$.

Step 2: Finding Alpha Interval

Since √(5) ≈ 2.236$\sqrt{5} \approx 2.236$, we have √(5)3 ≈ 0.745$\frac{\sqrt{5}}{3} \approx 0.745$. So α$\alpha$ must be strictly greater than 0.745$0.745$ (or less than -0.745$-0.745$).

Checking the given options: (1) 0.6 (No) (2) 0.8 (Yes, 0.8 > 0.745$0.8 > 0.745$) (3) 0.5 (No) (4) 0.7 (No)

Pattern Recognition

Geometrically, for a line not to meet a hyperbola, its slope must lie within a specific range determined by the asymptotes (m = ± b/a$m = \pm b/a$), and its c²$c^2$ must satisfy c² < a²m² - b²$c^2 < a^2m^2 - b^2$. Direct substitution to enforce D < 0$D < 0$ is purely mechanical and robust.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Core Logic

Given parabola y² = 12x 4a = 12 a = 3$y^2 = 12x \implies 4a = 12 \implies a = 3$.

Let the points be P₁(x₁, y₁) = (3t₁², 6t₁)$P_1(x_1, y_1) = (3t_1^2, 6t_1)$ and P₂(x₂, y₂) = (3t₂², 6t₂)$P_2(x_2, y_2) = (3t_2^2, 6t_2)$.

Since the chord subtends a right angle at the vertex (origin),

Step 1: Calculating the Expression

We need to evaluate x₁ x₂ - y₁ y₂$x_1 x_2 - y_1 y_2$:

Substitute t₁ t₂ = -4$t_1 t_2 = -4$:

= 9(16) + 144$= 9(16) + 144$

Pattern Recognition

Right angles subtended at the vertex by a chord on y² = 4ax$y^2 = 4ax$ instantly lock the parameter product to t₁ t₂ = -4$t_1 t_2 = -4$. Substitute this directly into any coordinate products required by the problem.

Chapter Mix

Class 11 Maths: Conic Sections

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