Let a circle of radius 4$4$ pass through the origin O$O$, the points A(-√(3)a, 0)$A(-\sqrt{3}a, 0)$ and B(0, -√(2)b)$B(0, -\sqrt{2}b)$, where a$a$ and b$b$ are real parameters and ab ≠ 0$ab \neq 0$. Then the locus of the centroid of Δ OAB$\Delta OAB$ is a circle of radius

Related Formula

Core Logic

Step 1: Calculate coordinates of B

For t₂ = -1/2$t_2 = -1/2$, point B$B$ is: x = 4(-1/2)² = 1$x = 4(-1/2)^2 = 1$ y = 8(-1/2) = -4$y = 8(-1/2) = -4$ So, B(1, -4)$B(1, -4)$.

Step 2: Section formula calculations (Two cases)

Point P(α, β)$P(\alpha, \beta)$ divides AB$AB$ in the ratio 5:2$5:2$. There are two possibilities depending on which end the ratio starts from.

Case 1: Ratio 5 from B to A (i.e. A is x₂$x_2$ and B is x₁$x_1$):

Sum: α + β = (154)/(7) = 22$\alpha + \beta = \frac{154}{7} = 22$.

Case 2: Ratio 5 from A to B (i.e. B is x₂$x_2$ and A is x₁$x_1$):

Sum: α + β = (49)/(7) = 7$\alpha + \beta = \frac{49}{7} = 7$.

Step 3: Minimum Value

Comparing the two possible sums, 7 < 22$7 < 22$. Thus, the minimum value is 7$7$.

Pattern Recognition

When a line segment is divided in a given ratio, 'internal division' inherently bears two solutions based on the orientation (from point A or point B). Always evaluate both cases when finding a minimum or maximum.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Core Logic

Step 1: Apply Slope Condition

Slope of bisector is 1$1$, so slope of PQ$PQ$ must be -1$-1$.

Step 2: Apply Midpoint Condition

Midpoint of PQ$PQ$ is ( (2 θ + α)/(2), (2 θ - 5α - 2)/(2) )$\left( \frac{2\cos\theta + \alpha}{2}, \frac{2\sin\theta - 5\alpha - 2}{2} \right)$. Substitute into x - y + 1 = 0$x - y + 1 = 0$:

Step 3: Eliminate alpha and Solve

From (1), 2α = - θ - θ - 1 α = (- θ - θ - 1)/(2)$2\alpha = -\sin\theta - \cos\theta - 1 \implies \alpha = \frac{-\sin\theta - \cos\theta - 1}{2}$. Substitute α$\alpha$ into (2):

Let's express in half angles:

So, (θ)/(2) = 0 θ = 1$\sin\frac{\theta}{2} = 0 \implies \cos\theta = 1$ or (θ)/(2) = 5 θ = (1 - ²(θ/2))/(1 + ²(θ/2)) = (1 - 25)/(1 + 25) = -(24)/(26) = -(12)/(13)$\tan\frac{\theta}{2} = 5 \implies \cos\theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - 25}{1 + 25} = -\frac{24}{26} = -\frac{12}{13}$.

Step 4: Final Calculation

The abscissa of P$P$ is 2 θ$2\cos\theta$. Values of abscissa are 2(1) = 2$2(1) = 2$ and 2(-(12)/(13)) = -(24)/(13)$2\left(-\frac{12}{13}\right) = -\frac{24}{13}$. Sum of abscissa values = 2 - (24)/(13) = (26 - 24)/(13) = (2)/(13)$= 2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$. We need 13 × (Sum) = 13 × (2)/(13) = 2$13 \times (\text{Sum}) = 13 \times \frac{2}{13} = 2$.

Pattern Recognition

Instead of finding the image of a generic circle point in a line, construct the reflection point Q$Q$ parameter, use slope logic (m₁m₂=-1$m_1m_2=-1$) and midpoint logic simultaneously to create a trigonometric linear equation.

Chapter Mix

Class 11 Maths: Circles Class 11 Maths: Straight Lines

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

where d$d$ is the distance between their centers.

Core Logic

Circle 1: (x + 1)² + (y + 4)² = r²$(x + 1)^2 + (y + 4)^2 = r^2$ Center C₁ = (-1, -4)$C_1 = (-1, -4)$ and Radius r₁ = r$r_1 = r$.

Circle 2: x² + y² - 4x - 2y - 4 = 0 (x - 2)² + (y - 1)² = 3²$x^2 + y^2 - 4x - 2y - 4 = 0 \implies (x - 2)^2 + (y - 1)^2 = 3^2$ Center C₂ = (2, 1)$C_2 = (2, 1)$ and Radius r₂ = 3$r_2 = 3$.

Step 1: Distance Between Centers

Distance d$d$ between C₁$C_1$ and C₂$C_2$:

Step 2: Applying the Intersection Condition

For two distinct intersection points:

Breaking this down into two inequalities:

• |r - 3| < √(34) -√(34) < r - 3 < √(34) 3 - √(34) < r < 3 + √(34)$|r - 3| < \sqrt{34} \implies -\sqrt{34} < r - 3 < \sqrt{34} \implies 3 - \sqrt{34} < r < 3 + \sqrt{34}$
• r + 3 > √(34) r > √(34) - 3$r + 3 > \sqrt{34} \implies r > \sqrt{34} - 3$

Since radius r > 0$r > 0$, taking the intersection of the conditions:

Thus, α = √(34) - 3$\alpha = \sqrt{34} - 3$ and β = √(34) + 3$\beta = \sqrt{34} + 3$.

Step 3: Calculating Final Product

Pattern Recognition

Intersection of two circles boils down to the fundamental triangle inequality relating the radii to the center distance: |r₁ - r₂| < d < r₁ + r₂$|r_1 - r_2| < d < r_1 + r_2$. Solving this naturally yields an interval (α, β)$(\alpha, \beta)$ formatted as a difference of squares upon multiplication.

Chapter Mix

Class 11 Maths: Circles

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