If α x + β y = 109$\alpha x + \beta y = 109$ is the equation of the chord of the ellipse (x²)/(9) +(y²)/(4) = 1$\frac{x^2}{9} +\frac{y^2}{4} = 1$, whose mid point is ((5)/(2),(1)/(2))$\left(\frac{5}{2},\frac{1}{2}\right)$, then α +β$\alpha +\beta$ is equal to

Related Formula

Core Logic

First, find the maximum value of f(t) = -(3)/(4) + 2t - t²$f(t) = -\frac{3}{4} + 2t - t^{2}$.

Completing the square or differentiating:

The maximum value is (1)/(4)$\frac{1}{4}$. Thus, eccentricity e = (1)/(4)$e = \frac{1}{4}$.

Step 1: Relating a and b

Given latus rectum is 30$30$:

Step 2: Solving for a and b

Substitute (2) into (1):

Since a ≠ 0$a \neq 0$, we have 15a - 240 = 0 a = 16$15a - 240 = 0 \implies a = 16$.

Then, b² = 15(16) = 240$b^{2} = 15(16) = 240$. So a² = 256$a^{2} = 256$.

Step 3: Final Calculation

We need to find a² + b²$a^{2} + b^{2}$:

Pattern Recognition

Whenever an ellipse's latus rectum and eccentricity are provided, it generates a standard system of two equations linking a$a$ and b²$b^{2}$. Solve for a$a$ first since b²$b^{2}$ is linear with respect to a$a$ via latus rectum.

Chapter Mix

Class 11 Maths: Ellipse Class 12 Maths: Application of Derivatives

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Core Logic

Take a general parametric point P$P$ on the parabola x² = 4y$x^2 = 4y$, which is P(2t, t²)$P(2t, t^2)$.

We find the mirror image Q(h, k)$Q(h, k)$ of P$P$ with respect to the line x - y - 1 = 0$x - y - 1 = 0$.

Step 1: Finding the Image Coordinates

Solving for h$h$:

Solving for k$k$:

Step 2: Eliminating the Parameter

From k = 2t - 1$k = 2t - 1$, we get t = (k + 1)/(2)$t = \frac{k + 1}{2}$.

Substitute t$t$ into h$h$:

Step 3: Finding Target Values

Replacing (h, k)$(h, k)$ with (x, y)$(x, y)$, the image parabola is:

Comparing this with (y + α)² = b(x - c)$(y + \alpha)^2 = b(x - c)$: α = 1, b = 4, c = 1$\alpha = 1, b = 4, c = 1$

Pattern Recognition

To find the image of a conic section across a linear axis, it is almost always computationally cleaner to reflect its general parametric point rather than manipulating the implicit Cartesian equation through coordinate transformations.

Chapter Mix

Class 11 Maths: Parabola Class 11 Maths: Straight Lines

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Core Logic

Let the point P(h, k)$P(h, k)$ lie on x² + y² = 4$x^2 + y^2 = 4$. Using parametric coordinates:

Let the target point be Q(x, y)$Q(x, y)$:

y = 3k + 2 = 3(2 θ) + 2 = 6 θ + 2$$y = 3k + 2 = 3(2\sin\theta) + 2 = 6\sin\theta + 2$$ (Wait, PDF states 3k+2$3k+2$, parametric solution in PDF says 6 θ + 3$6\sin\theta + 3$. Checking exact text: "(2h + 1, 3k + 2)$(2h + 1, 3k + 2)$ lie on an ellipse..." but solution uses "6 θ + 3$6\sin\theta + 3$". If the question meant 3k+3$3k+3$, that would be a typo in the question paper. However, 3(2 θ)+2 = 6 θ+2$3(2\sin\theta)+2 = 6\sin\theta+2$. This implies (y-2)/(6) = θ$\frac{y-2}{6} = \sin\theta$. Either way, the denominators a$a$ and b$b$ of the resulting ellipse remain 4 and 6, so eccentricity is invariant to the constant offset).

Step 1: Finding the Locus

Isolating θ$\cos\theta$ and θ$\sin\theta$:

Using ²θ + ²θ = 1$\sin^2\theta + \cos^2\theta = 1$:

This is the equation of an ellipse where a = 4$a = 4$ and b = 6$b = 6$ (since b > a$b > a$, it's a vertical ellipse).

Step 2: Calculating Eccentricity

For a vertical ellipse (b > a$b > a$), eccentricity is:

Step 3: Finding Final Target

We need the value of (5)/(e²)$\frac{5}{e^2}$:

Pattern Recognition

Affine transformations (ax+b, cy+d)$(ax+b, cy+d)$ applied to a circle's locus purely stretch its semi-axes to the respective scaling constants (a$a$ and c$c$). Translational constants (b$b$ and d$d$) shift the center but do not affect the eccentricity.

Chapter Mix

Class 11 Maths: Ellipse Class 11 Maths: Circles

Core Logic

For the hyperbola (x²)/(8) - (y²)/(8 ²θ) = 1$\frac{x^2}{8} - \frac{y^2}{8\cos^2\theta} = 1$: a² = 8, b² = 8 ²θ$a^2 = 8, b^2 = 8\cos^2\theta$ Eccentricity e₁ = √(1 + (b²)/(a²)) = √(1 + (8 ²θ)/(8)) = √(1 + ²θ)$e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{8\cos^2\theta}{8}} = \sqrt{1 + \cos^2\theta}$. Latus rectum ₁ = (2b²)/(a) = 2(8 ²θ)2√(2) = 4√(2) ²θ$\ell_1 = \frac{2b^2}{a} = \frac{2(8\cos^2\theta)}{2\sqrt{2}} = 4\sqrt{2}\cos^2\theta$.

For the ellipse (x²)/(6 ²θ) + (y²)/(6) = 1$\frac{x^2}{6\cos^2\theta} + \frac{y^2}{6} = 1$: Here a² = 6 ²θ, b² = 6$a^2 = 6\cos^2\theta, b^2 = 6$. Since θ in (0, π/2)$\theta \in (0, \pi/2)$, 6 > 6 ²θ$6 > 6\cos^2\theta$, so the major axis is along the y-axis. Eccentricity e₂ = √(1 - (a²)/(b²)) = √(1 - (6 ²θ)/(6)) = √(1 - ²θ) = θ$e_2 = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{6\cos^2\theta}{6}} = \sqrt{1 - \cos^2\theta} = \sin\theta$. Latus rectum ₂ = (2a²)/(b) = 2(6 ²θ)√(6) = 2√(6) ²θ$\ell_2 = \frac{2a^2}{b} = \frac{2(6\cos^2\theta)}{\sqrt{6}} = 2\sqrt{6}\cos^2\theta$.

Step 1: Solve for Theta

Given relation: e₁² = e₂² ( ²θ + 1)$e_1^2 = e_2^2 (\sec^2\theta + 1)$

Replace ²θ$\sin^2\theta$ with 1 - ²θ$1 - \cos^2\theta$:

Factorizing:

Since ²θ + 1 ≠ 0$\cos^2\theta + 1 \neq 0$, we have 2 ²θ = 1 ²θ = (1)/(2)$2\cos^2\theta = 1 \implies \cos^2\theta = \frac{1}{2}$. Since θ in (0, π/2)$\theta \in (0, \pi/2)$, θ = (π)/(4)$\theta = \frac{\pi}{4}$.

Step 2: Evaluate All Variables

For θ = π/4$\theta = \pi/4$: e₁ = √(1 + 1/2) = √(3/2)$e_1 = \sqrt{1 + 1/2} = \sqrt{3/2}$ e₂ = (π/4) = 1/√(2)$e_2 = \sin(\pi/4) = 1/\sqrt{2}$ ₁ = 4√(2)(1/2) = 2√(2)$\ell_1 = 4\sqrt{2}(1/2) = 2\sqrt{2}$ ₂ = 2√(6)(1/2) = √(6)$\ell_2 = 2\sqrt{6}(1/2) = \sqrt{6}$ ²θ = 1$\tan^2\theta = 1$

Step 3: Final Calculation

Evaluate the target expression:

Numerator: 2√(12) = 4√(3)$2\sqrt{12} = 4\sqrt{3}$ Denominator: √(3/4) = √(3)2$\sqrt{3/4} = \frac{\sqrt{3}}{2}$ Result = 4√(3) √(3)2 = 8$\frac{4\sqrt{3}}{\frac{\sqrt{3}}{2}} = 8$

Chapter Mix

Class 11 Mathematics: Conic Sections

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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