Let C be the circle x² + (y - 1)² = 2$x^2 + (y - 1)^2 = 2$, E₁$E_1$ and E₂$E_2$ be two ellipses whose centres lie at the origin and major axes lie on the x-axis and y-axis respectively. Let the straight line x + y = 3$x + y = 3$ touch the curves C, E₁$E_1$ and E₂$E_2$ at P(x₁, y₁)$P(x_1, y_1)$, Q(x₂, y₂)$Q(x_2, y_2)$ and R(x₃, y₃)$R(x_3, y_3)$ respectively. Given that P is the mid-point of the line segment QR and PQ = 2√(2)3$PQ = \frac{2\sqrt{2}}{3}$, the value of 9(x₁y₁ + x₂y₂ + x₃y₃)$9(x_1y_1 + x_2y_2 + x_3y_3)$ is equal to

Related Formula

Core Logic

Focus of y² = 8x$y^2 = 8x$ is A(2, 0)$A(2, 0)$ since 4a = 8 ⇒ a=2$4a = 8 \Rightarrow a=2$. Let points on parabola be B(2t₁², 4t₁)$B(2t_1^2, 4t_1)$ and C(2t₂², 4t₂)$C(2t_2^2, 4t_2)$. The centroid of Δ ABC$\Delta ABC$ is given as ((7)/(3), (4)/(3))$\left(\frac{7}{3}, \frac{4}{3}\right)$. Equating coordinates:

Execution

Square the sum equation:

Find (t₁ - t₂)²$(t_1 - t_2)^2$:

Calculate distance squared for BC$BC$:

Pattern Recognition

Using parametric coordinates (at², 2at)$(at^2, 2at)$ systematically reduces algebraic complexity when determining intersections or triangle properties on a parabola.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Core Logic

From the given coordinates on the major axis (y = -2$y = -2$): Center C(1, -2)$C(1, -2)$, Focus F₁(3, -2)$F_1(3, -2)$, Vertex A₁(5, -2)$A_1(5, -2)$. Distance from center to vertex, CA₁ = a = 5 - 1 = 4$CA_1 = a = 5 - 1 = 4$. Distance from center to focus, CF₁ = ae = 3 - 1 = 2$CF_1 = ae = 3 - 1 = 2$.

Execution

Calculate eccentricity e$e$:

Use alternate formula for Latus Rectum:

Pattern Recognition

Aligning focus, center, and vertex along a constant y-axis implies a standard shifted ellipse where absolute differences in x-coordinates yield standard parameters (a$a$ and ae$ae$) directly.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Core Logic

For Ellipse E: (x²)/(144) + (y²)/(169) = 1$E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ a² = 144, b² = 169$a^2 = 144, b^2 = 169$. Since b > a$b > a$, the major axis is along the y-axis. Eccentricity e' = √(1 - (144)/(169)) = √((25)/(169)) = (5)/(13)$e' = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$. Foci of ellipse = (0, ± be') = (0, ± 13 × (5)/(13)) = (0, ± 5)$(0, \pm be') = (0, \pm 13 \times \frac{5}{13}) = (0, \pm 5)$.

Execution

For Hyperbola H: (y²)/(λ²) - (x²)/(16) = 1$H: \frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$ Foci of conjugate hyperbola are (0, ± λ e)$(0, \pm \lambda e)$. Equating foci: λ e = 5$\lambda e = 5$.

Eccentricity of hyperbola, e = (5)/(3)$e = \frac{5}{3}$. Length of latus rectum of hyperbola, L = (2(16))/(λ) = (32)/(3)$L = \frac{2(16)}{\lambda} = \frac{32}{3}$.

Calculate 24(e + L)$24(e + L)$:

Pattern Recognition

Confocal conics usually align along the same major axis. Notice the -1$-1$ on the RHS of the hyperbola equation indicates a conjugate hyperbola orienting it vertically to match the b>a$b>a$ ellipse.

Chapter Mix

Class 11 Maths: Conic Sections

Core Logic

Intersection of circle with x-axis provides A(2,0)$A(2,0)$ and B(-2,0)$B(-2,0)$. Let the point of intersection of AQ$AQ$ and BP$BP$ be R(h, k)$R(h, k)$. Since R$R$ lies on BP$BP$, the slope mBR = mBP$m_{BR} = m_{BP}$:

Since R$R$ lies on AQ$AQ$, the slope mAR = mAQ$m_{AR} = m_{AQ}$:

Execution

We are given α - β = (π)/(2) ⇒ (α)/(2) - (β)/(2) = (π)/(4)$\alpha - \beta = \frac{\pi}{2} \Rightarrow \frac{\alpha}{2} - \frac{\beta}{2} = \frac{\pi}{4}$. Applying the (A-B)$\tan(A-B)$ formula:

Substitute the slope relations: (α)/(2) = (k)/(h+2)$\tan\frac{\alpha}{2} = \frac{k}{h+2}$ (β)/(2) = -(h-2)/(k)$\tan\frac{\beta}{2} = -\frac{h-2}{k}$ (since - (β)/(2) = (k)/(h-2)$-\cot\frac{\beta}{2} = \frac{k}{h-2}$)

1 = ((k)/(h+2) + (h-2)/(k))/(1 + ((k)/(h+2))((2-h)/(k)))$1 = \frac{\frac{k}{h+2} + \frac{h-2}{k}}{1 + \left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right)}$ 1 = k² + h² - 4(k(h+2))/(k) · (k(h+2) - k(2-h))/(k(h+2)) wait, clear denominator$1 = \frac{k^2 + h^2 - 4}{\frac{k(h+2)}{k} \cdot \frac{k(h+2) - k(2-h)}{k(h+2)} \dots \text{ wait, clear denominator}}$ 1 = (k² + h² - 4)/(k(h+2) + k(2-h)) × k(h+2)$1 = \frac{k^2 + h^2 - 4}{k(h+2) + k(2-h)} \times k(h+2) \dots$ The denominator simplifies to:

Numerator is (k² + h² - 4)/(k(h+2))$\frac{k^2 + h^2 - 4}{k(h+2)}$. So the expression simplifies to:

Locus of R$R$ is x² + y² - 4y - 4 = 0$x^2 + y^2 - 4y - 4 = 0$.

Pattern Recognition

Connecting chords from extreme diameter vertices to points whose parametric angles differ by π/2$\pi/2$ reliably generates perpendicular-like slope products or standard tangent angle identities, mapping directly to a circular locus.

Chapter Mix

Class 11 Maths: Circles

Related Formula

Core Logic

We set up an algebraic equation relating b$b$, a$a$, and e$e$ from the given geometric condition, then substitute it into the eccentricity identity.

Step 1: Set up the geometric relation

Given that 2b = (1)/(4) (2ae)$2b = \frac{1}{4} (2ae)$:

Square both sides:

Step 2: Solve for eccentricity

Using the eccentricity relation:

Pattern Recognition

Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing b/a$b/a$ as a function of e$e$ allows direct solving of the eccentricity.

Chapter Mix

Class 11 Mathematics: Conic Sections