Let ABCD be a trapezium whose vertices lie on the parabola y² = 4x$y^2 = 4x$. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length (25)/(4)$\frac{25}{4}$ and it passes through the point (1,0)$(1,0)$, then the area of ABCD is:

Related Formula

For a horizontal hyperbola centered at (h,k)$(h,k)$:

• Foci: (h ± ae, k)$(h \pm ae, k)$
• Eccentricity relation: b² = a²(e² - 1) = a² e² - a²$b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2$

Core Logic

Foci are S₁ = (4,2)$S_1 = (4,2)$ and S₂ = (8,2)$S_2 = (8,2)$.

• Center C(h,k)$C(h,k)$ is the midpoint:

• Distance between foci:

Thus, b² = 4 - a²$b^2 = 4 - a^2$.

Step 1: Expanding standard equation

The equation is:

Comparing with 3x² - y² - α x + β y + γ = 0$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$, the ratio of coefficients of x²$x^2$ and y²$y^2$ is (3)/(-1) = -3$\frac{3}{-1} = -3$:

Thus, b² = 4 - 1 = 3$b^2 = 4 - 1 = 3$.

Step 2: Finding values of coefficients α, β, γ$\alpha, \beta, \gamma$

Substituting a² = 1$a^2 = 1$ back into standard form equation:

Comparing coefficients:

• α = 36$\alpha = 36$
• β = 4$\beta = 4$
• γ = 101$\gamma = 101$

Pattern Recognition

Symmetric focal coordinates (y=2$y=2$) indicate the hyperbola is horizontal. Identifying coordinates of the center (6,2)$(6,2)$ quickly and using coefficient ratio comparison restricts parameters immediately without requiring complex algebraic systems.

Chapter Mix

Class 11 Conic Sections

Related Formula

By the definition of a parabola, the distance from any point (x, y)$(x, y)$ on the curve to the focus (xf, yf)$(x_f, y_f)$ equals its perpendicular distance to the directrix line Ax + By + C = 0$Ax + By + C = 0$:

Core Logic

Substituting the focus (-2, 1)$(-2, 1)$ and directrix 2x + y + 2 = 0$2x + y + 2 = 0$ into the definition equation:

Step 1: Substitute the Given Abscissa

Step 2: Find the Sum of Ordinates

The ordinates y₁$y_1$ and y₂$y_2$ are the roots of the quadratic equation 4y² - 6y + 1 = 0$4y^2 - 6y + 1 = 0$. The sum of the ordinates is:

Pattern Recognition

Notice how evaluating the intersection layout directly simplifies when the substitution value matches the coordinate of the focus, converting the entire quadratic horizontal layout component to 0 immediately.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For a standard hyperbola:

• Focus positions are (± ae, 0)$(\pm ae, 0)$.
• Length of semi-latus rectum is (b²)/(a)$\frac{b^2}{a}$.
• Eccentricity identity linkage: b² = a²(e² - 1) a²e² = a² + b²$b^2 = a^2(e^2 - 1) \implies a^2e^2 = a^2 + b^2$.

Core Logic

Given focus F₁ ≡ (-ae, 0) ≡ P(-3, 0)$F_1 \equiv (-ae, 0) \equiv P(-3, 0)$, so ae = 3$ae = 3$. The other focus is F₂ ≡ (ae, 0) ≡ (3, 0)$F_2 \equiv (ae, 0) \equiv (3, 0)$.

The latus rectum passes vertically through F₂$F_2$, with endpoints L₁(ae, (b²)/(a))$L_1\left(ae, \frac{b^2}{a}\right)$ and L₂(ae, -(b²)/(a))$L_2\left(ae, -\frac{b^2}{a}\right)$. This segment subtends a right angle at P(-ae, 0)$P(-ae, 0)$. By symmetry, the top half angle at P$P$ must be exactly 45^°$45^\circ$.

Step 1: Set Up Slope Relationship

Step 2: Solve the Quadratic Excentricity Equation

Substitute ae = 3$ae = 3$ and b² = 6a$b^2 = 6a$ into the eccentricity identity a²e² = a² + b²$a^2e^2 = a^2 + b^2$:

Solving for a$a$ using the quadratic formula (taking the positive root since a > 0$a > 0$):

Step 3: Evaluate product and sum coefficients

Now compute a²b²$a^2b^2$:

Matching with α√(2) - β$\alpha\sqrt{2} - \beta$ gives:

Calculate the final required sum:

Pattern Recognition

Recognizing that the right angle subtended at the opposite focus implies a perfect (45^°)$\tan(45^\circ)$ right triangle instantly yields the key linear constraint b² = 2a(ae)$b^2 = 2a(ae)$, avoiding the need for lengthy distance-formula tracking.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

Core Logic

Extract the coordinate center of the target circle, substitute it directly to locate the missing parameter p$p$, and resolve eccentricity metrics.

Step 1: Extract Circle Metric Values

For circle x² + y² - 2x - 4y - 11 = 0$x^{2} + y^{2} - 2x - 4y - 11 = 0$:

Step 2: Standardize Ellipse Formulation

Ellipse passes through point C(1,2)$C(1,2)$:

Standard model form: (x²)/(4/3) + (y²)/(16) = 1$\frac{x^2}{4/3} + \frac{y^2}{16} = 1$ (b > a$b > a$, vertical configuration axis).

Step 3: Evaluate Product Chain

Focal distance elements at ordinate coordinate height k=2$k=2$ are bounded by b ± ek$b \pm ek$:

Target evaluation expression response string:

Pattern Recognition

Pay attention to whether b > a$b > a$ or a > b$a > b$ when analyzing ellipse forms. Focal distance definitions swap directions immediately across major horizontal/vertical configurations.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Related Formula

Core Logic

Establish the tangent vector expression at the parabola intersection mark. Since this path line functions as a shared contact tangent boundaries sheet for the circular arc, impose radius equations.

Step 1: Derive Shared Parabola Tangent Line

Tangent line profile for y² = 9x$y^2 = 9x$ at coordinate indicator (4,6)$(4,6)$:

Step 2: Build Geometric Metric Connections

Circle touches axis at (a,0)$(a,0)$, mapping coordinates center directly to C(a,r)$C(a,r)$. Perpendicular boundary constraint steps require:

Step 3: Solve for Radius Matrix Bounds

Enforce circle equation intersection constraint profile (x-a)² + (y-r)² = r²$(x-a)^2 + (y-r)^2 = r^2$ at point (4,6)$(4,6)$:

Evaluating the target systems from structural logic tracks rejects positive value parameters, providing:

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Pattern Recognition

Shared tangent elements connect independent conic fields. Locating circular center boundaries using axial coordinate tracking simplifies secondary equations.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles