If the equation of the parabola with vertex V((3)/(2),3)$V\left(\frac{3}{2},3\right)$ and the directrix x+2y=0$x+2y=0$ is α x²+β y²-γ xy-30x-60y+225=0,$\alpha x^{2}+\beta y^{2}-\gamma xy-30x-60y+225=0,$ then α+β+γ$\alpha+\beta+\gamma$ is equal to:

Related Formula

Latus Rectum = 2b²a$$\text{Latus Rectum } = \frac{2b^{2}}{a}$$ e² = 1 - b²a² = a²-b²a²$$e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{a^{2}-b^{2}}{a^{2}}$$

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:

f(t) = -(t² - 2t + (3)/(4)) = -((t-1)² - 1 + (3)/(4)) = (1)/(4) - (t-1)²$$f(t) = -\left(t^{2} - 2t + \frac{3}{4}\right) = -\left((t-1)^{2} - 1 + \frac{3}{4}\right) = \frac{1}{4} - (t-1)^{2}$$

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

e² = (1)/(16) a² - b²a² = (1)/(16) (1)$$e^{2} = \frac{1}{16} \implies \frac{a^{2} - b^{2}}{a^{2}} = \frac{1}{16} \quad \dots (1)$$

Step 1: Relating a and b

Given latus rectum is 30$30$:

2b²a = 30 b² = 15a (2)$$\frac{2b^{2}}{a} = 30 \implies b^{2} = 15a \quad \dots (2)$$

Step 2: Solving for a and b

Substitute (2) into (1):

16(a² - 15a) = a²$$16(a^{2} - 15a) = a^{2}$$ 15a² - 240a = 0$$15a^{2} - 240a = 0$$

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$a^{2} + b^{2}$:

a² + b² = 256 + 240 = 496$$a^{2} + b^{2} = 256 + 240 = 496$$

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

Related Formula

Image of point (x₁, y₁) in line ax + by + c = 0 is given by:$$\text{Image of point } (x_1, y_1) \text{ in line } ax + by + c = 0 \text{ is given by:}$$ (x - x₁)/(a) = (y - y₁)/(b) = -2 (ax₁ + by₁ + c)/(a² + b²)$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$$

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

(h - 2t)/(1) = (k - t²)/(-1) = -2 (2t - t² - 1)/(1² + (-1)²)$$\frac{h - 2t}{1} = \frac{k - t^2}{-1} = -2 \frac{2t - t^2 - 1}{1^2 + (-1)^2}$$ (h - 2t)/(1) = (k - t²)/(-1) = -(2t - t² - 1) = t² - 2t + 1$$\frac{h - 2t}{1} = \frac{k - t^2}{-1} = -(2t - t^2 - 1) = t^2 - 2t + 1$$

Solving for h$h$:

h - 2t = t² - 2t + 1 h = t² + 1$$h - 2t = t^2 - 2t + 1 \implies h = t^2 + 1$$

Solving for k$k$:

k - t² = -(t² - 2t + 1) = -t² + 2t - 1 k = 2t - 1$$k - t^2 = -(t^2 - 2t + 1) = -t^2 + 2t - 1 \implies k = 2t - 1$$

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$h$:

h = ((k + 1)/(2))² + 1$$h = \left(\frac{k + 1}{2}\right)^2 + 1$$ h - 1 = ((k + 1)²)/(4)$$h - 1 = \frac{(k + 1)^2}{4}$$ (k + 1)² = 4(h - 1)$$(k + 1)^2 = 4(h - 1)$$

Step 3: Finding Target Values

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

(y + 1)² = 4(x - 1)$$(y + 1)^2 = 4(x - 1)$$

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

a + b + c = 1 + 4 + 1 = 6$$a + b + c = 1 + 4 + 1 = 6$$

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

Related Formula

Parametric form of a circle x² + y² = r²: (r θ, r θ)$$\text{Parametric form of a circle } x^2 + y^2 = r^2: (r\cos\theta, r\sin\theta)$$ Eccentricity of an ellipse: e² = 1 - (b²)/(a²) (where a > b)$$\text{Eccentricity of an ellipse: } e^2 = 1 - \frac{b^2}{a^2} \text{ (where } a > b)$$

Core Logic

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

h = 2 θ, k = 2 θ$$h = 2\cos\theta, \quad k = 2\sin\theta$$

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

x = 2h + 1 = 2(2 θ) + 1 = 4 θ + 1$$x = 2h + 1 = 2(2\cos\theta) + 1 = 4\cos\theta + 1$$

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$\sin\theta$:

θ = (x - 1)/(4)$$\cos\theta = \frac{x - 1}{4}$$ θ = (y - 2)/(6) (or (y-3)/(6) per the solution)$$\sin\theta = \frac{y - 2}{6} \quad \text{(or } \frac{y-3}{6} \text{ per the solution)}$$

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

( (x - 1)/(4) )² + ( (y - 2)/(6) )² = 1$$\left( \frac{x - 1}{4} \right)^2 + \left( \frac{y - 2}{6} \right)^2 = 1$$

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:

e² = 1 - (a²)/(b²) = 1 - (4²)/(6²) = 1 - (16)/(36)$$e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{4^2}{6^2} = 1 - \frac{16}{36}$$ e² = (36 - 16)/(36) = (20)/(36) = (5)/(9)$$e^2 = \frac{36 - 16}{36} = \frac{20}{36} = \frac{5}{9}$$

Step 3: Finding Final Target

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

(5)/(e²) = (5)/((5)/(9)) = 9$$\frac{5}{e^2} = \frac{5}{\frac{5}{9}} = 9$$

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

Equation of circle C₂$C_{2}$ with diameter along y=x$y=x$ passing through origin and having length 10. Wait, C₁$C_1$ has diameter 10. The chord y=x$y=x$ passes through (0,0)$(0,0)$. For the chord to be a diameter of C₂$C_2$, the points of intersection with C₁$C_1$ must form the diameter. The center of C₂$C_2$ is the midpoint of the chord. Let the ends of the chord be (0,0)$(0,0)$ and (5,5)$(5,5)$ (since length is √(50)$\sqrt{50}$? Wait, the problem implies the chord of C₁$C_1$ is y=x$y=x$. If C₁$C_1$ is a circle in x ≥ 0$x \geq 0$ of diameter 10 through origin. Center of C₂$C_2$ lies on the chord y=x$y=x$. The equation of circle C₂$C_{2}$ is:

x² + y² - 5x - 5y = 0$$x^2 + y^2 - 5x - 5y = 0$$

Its center is N((5)/(2), (5)/(2))$N\left(\frac{5}{2}, \frac{5}{2}\right)$.

Step 1: Find Farthest Chord

We need the chord of C₂$C_2$ passing through B(2, 3)$B(2, 3)$ which is farthest from the center N((5)/(2), (5)/(2))$N\left(\frac{5}{2}, \frac{5}{2}\right)$. The farthest chord passing through a given point is always perpendicular to the line joining the center to that point. Slope of line NB$NB$:

mNB = (3 - (5)/(2))/(2 - (5)/(2)) = ((1)/(2))/(-(1)/(2)) = -1$$m_{NB} = \frac{3 - \frac{5}{2}}{2 - \frac{5}{2}} = \frac{\frac{1}{2}}{-\frac{1}{2}} = -1$$

Step 2: Chord Equation

The slope of the required chord is perpendicular to mNB$m_{NB}$:

Slope of required chord = 1$$\text{Slope of required chord } = 1$$

Equation of the required chord passing through (2,3)$(2,3)$:

y - 3 = 1(x - 2)$$y - 3 = 1(x - 2)$$

x - y + 1 = 0$x - y + 1 = 0$ Comparing this with x + ay + b = 0$x + ay + b = 0$, we get:

a = -1, b = 1$$a = -1, \quad b = 1$$

Step 3: Final Calculation

a - b = -1 - 1 = -2$$a - b = -1 - 1 = -2$$

Pattern Recognition

The chord of a circle passing through a given internal point that is FARTHEST from the center is exactly the chord that is PERPENDICULAR to the radius (or line segment) connecting the center to that internal point.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

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)$e_1^2 = e_2^2 (\sec^2\theta + 1)$

1 + ²θ = ²θ (1 + (1)/( ²θ))$$ 1 + \cos^2\theta = \sin^2\theta \left(1 + \frac{1}{\cos^2\theta}\right) $$ 1 + ²θ = ²θ + ²θ$$ 1 + \cos^2\theta = \sin^2\theta + \tan^2\theta $$

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

1 + ²θ = 1 - ²θ + ²θ$$ 1 + \cos^2\theta = 1 - \cos^2\theta + \tan^2\theta $$ 2 ²θ = ²θ = ( ²θ)/( ²θ)$$ 2\cos^2\theta = \tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta} $$ 2 ⁴θ = 1 - ²θ 2 ⁴θ + ²θ - 1 = 0$$ 2\cos^4\theta = 1 - \cos^2\theta \implies 2\cos^4\theta + \cos^2\theta - 1 = 0 $$

Factorizing:

(2 ²θ - 1)( ²θ + 1) = 0$$ (2\cos^2\theta - 1)(\cos^2\theta + 1) = 0 $$

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:

(( ₁ ₂)/(e₁ e₂)) ² θ = 2√(2) · √(6)√(3/2) · 1/√(2) · 1$$ \left(\frac{\ell_1 \ell_2}{e_1 e_2}\right) \tan^2 \theta = \frac{2\sqrt{2} \cdot \sqrt{6}}{\sqrt{3/2} \cdot 1/\sqrt{2}} \cdot 1 $$

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