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

Shortest distance between symmetric profiles: The minimal spacing normal line runs completely perpendicular to the mutual line of symmetry y=x$y=x$.

Core Logic

The given curve equations reflect symmetry across line y=x$y=x$[cite: 1382, 1383]. The tangent slope at the closest matching locations must run parallel to this mirror path [cite: 1403]: dydx = 1$$\frac{\mathrm{d}y}{\mathrm{d}x} = 1$$ [cite: 1403]

Differentiate curve equation y = x² + 2$y = x^2 + 2$ [cite: 1404]: dydx = 2x = 1 x = (1)/(2)$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x = 1 \implies x = \frac{1}{2}$$ [cite: 1405, 1406] Substitute back to get y-coordinate [cite: 1406]: y = ((1)/(2))² + 2 = (9)/(4) B((1)/(2), (9)/(4))$$y = \left(\frac{1}{2}\right)^2 + 2 = \frac{9}{4} \implies B\left(\frac{1}{2}, \frac{9}{4}\right)$$ [cite: 1406, 1407]

By mirror symmetry, the corresponding point on the other parabola is [cite: 1407]: A((9)/(4), (1)/(2))$$A\left(\frac{9}{4}, \frac{1}{2}\right)$$ [cite: 1407]

Step 1: Calculating distance and circle radius

Evaluate chord distance AB$AB$ using standard metrics [cite: 1407]: AB = √(((9)/(4) - (1)/(2))² + ((1)/(2) - (9)/(4))²) = √(2 · ((7)/(4))²) = 7√(2)4$$AB = \sqrt{\left(\frac{9}{4} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{9}{4}\right)^2} = \sqrt{2 \cdot \left(\frac{7}{4}\right)^2} = \frac{7\sqrt{2}}{4}$$ [cite: 1407, 1408]

The diameter of the smallest circle spanning between these touching curves equals distance AB$AB$ [cite: 1408]. Radius = (AB)/(2) = 7√(2)8$$\text{Radius} = \frac{AB}{2} = \frac{7\sqrt{2}}{8}$$ [cite: 1408]

Pattern Recognition

Mutually inverse conic curves track symmetric footprints. Their closest distance segments always align perfectly perpendicular to the main baseline axis line y=x$y=x$.

Chapter Mix

Class 11 Mathematics: Conic Sections (Parabola)

Related Formula

For hyperbola conics:

• Focal distances product: PS₁ · PS₂ = |a²e² - x²|$PS_1 \cdot PS_2 = |a^2e^2 - x^2|$
• Point lying on curve constraint verification properties.

Core Logic

Since point P(4, 2√(3))$P(4, 2\sqrt{3})$ resides directly on hyperbola curve structure [cite: 1445]: (16)/(a²) - (12)/(b²) = 1 16b² - 12a² = a²b²$$\frac{16}{a^2} - \frac{12}{b^2} = 1 \implies 16b^2 - 12a^2 = a^2b^2$$ [cite: 1446, 1448]

Using focal coordinate geometric spacing properties [cite: 1445, 1451]: PS₁ = ae - 4, PS₂ = ae + 4 PS₁ · PS₂ = a²e² - 16 = 32$$PS_1 = ae - 4, \quad PS_2 = ae + 4 \implies PS_1 \cdot PS_2 = a^2e^2 - 16 = 32$$ [cite: 1445, 1451] a²e² = 48 a² + b² = 48$$a^2e^2 = 48 \implies a^2 + b^2 = 48$$ [cite: 1452, 1453]

Step 1: Solving axis components values

Substitute b² = 48 - a²$b^2 = 48 - a^2$ back into original parameter product template [cite: 1448]:

Testing parameters [cite: 1454]: From relation b² - a² = 4$b^2 - a^2 = 4$ [cite: 1454], we resolve the dimensions [cite: 1457, 1458]: a² = 8, b² = 12$$a^2 = 8, \quad b^2 = 12$$ [cite: 1457, 1458]

Step 2: Total Calculation

Length formulas for targeted metrics [cite: 1459]:

Final Metric Total = p² + q² = 48 + 72 = 120$$\text{Final Metric Total} = p^2 + q^2 = 48 + 72 = 120$$ [cite: 1459, 1460]

Pattern Recognition

Focal calculations relative to specific points simplify elegantly under eccentricity conversions. Solving quadratic frames sequentially ensures structural accuracy.

Chapter Mix

Class 11 Mathematics: Conic Sections (Hyperbola)

Core Logic

The circle C$C$ has its center at the origin and passes through the foci F₁(-ae, 0)$F_1(-ae, 0)$ and F₂(ae, 0)$F_2(ae, 0)$. This means F₁F₂$F_1F_2$ is the diameter of the circle. Any point P$P$ lying on this circle satisfies the property that the angle subtended by the diameter is a right angle:

Step 1: Using the Area and Ellipse Definition

Since PF₁F₂$\triangle PF_1F_2$ is a right-angled triangle at P$P$:

By the definition of an ellipse, the sum of the focal distances to any point on the curve is equal to the length of the major axis (2a = 17$2a = 17$):

Step 2: Calculating Distance between Foci

Applying Pythagoras' theorem in right-angled PF₁F₂$\triangle PF_1F_2$:

Substitute the known values from our equations block:

Therefore, the distance between the foci is 13$13$.

Pattern Recognition

Whenever a circle is circumscribed around the foci of an ellipse, remember Thales' theorem: any intersection point with the ellipse forms a right triangle with the focal diameter, linking focal properties directly to Pythagoras.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For any point P$P$ on a parabola, its distance to the focus S$S$ equals its perpendicular distance to the directrix line M$M$:

PS = PM$PS = PM$

Core Logic

The axis line is y = x$y = x$. The vertex lies along this line at a distance of √(2)$\sqrt{2}$ from the origin. Since it's in the first quadrant, its coordinates are (1,1)$(1,1)$. The focus also lies along y=x$y=x$ at a distance of 2√(2)$2\sqrt{2}$ from the origin, which gives coordinates (2,2)$(2,2)$.

Step 1: Finding the Equation of the Directrix

The distance from the vertex to the focus is a = √((2-1)² + (2-1)²) = √(2)$a = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$. The directrix is perpendicular to the axis line y = x$y = x$ (slope = 1$1$), so the slope of the directrix is -1$-1$. The directrix is located at a distance a = √(2)$a = \sqrt{2}$ behind the vertex, which brings it exactly to the origin (0,0)$(0,0)$.

Therefore, the equation of the directrix line is:

Step 2: Utilizing the Focus-Directrix Property

Let the point P(1,k)$P(1,k)$ lie on the parabola. Applying PS = PM$PS = PM$:

Squaring both sides:

Factoring the quadratic equations:

Pattern Recognition

When a vertex and focus both sit perfectly on a symmetric line like y=x$y=x$, notice that the foot of the directrix often lands on a clean coordinate intersection (like the origin here), heavily simplifying geometric distance steps.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For a point P(x₁, y₁)$P(x_1, y_1)$ on a hyperbola branch, the focal distances are ex₁ + a$ex_1 + a$ and ex₁ - a$ex_1 - a$. Their sum is 2ex₁$2ex_1$, and their product is e²x₁² - a²$e^2x_1^2 - a^2$.

Core Logic

Given the point P(4,3)$P(4,3)$, the x-coordinate is x₁ = 4$x_1 = 4$. The sum of focal distances is:

Using the eccentricity relation b² = a²(e² - 1)$b^2 = a^2(e^2 - 1)$:

Step 1: Finding the Ellipse Parameters

Since P(4,3)$P(4,3)$ lies on the hyperbola (x²)/(a²) - (y²)/(b²) = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

Now calculate b²$b^2$:

Step 2: Calculating l^2 and m

The length of the latus rectum l$l$ is given by l = (2b²)/(a)$l = \frac{2b^2}{a}$:

The product of focal distances m$m$ is:

6m = 145$6m = 145$

Step 3: Final Computation

Evaluating the targeted expression:

Pattern Recognition

Using focal property formulas directly (2ex₁$2ex_1$ for sum and e²x₁² - a²$e^2x_1^2 - a^2$ for product) avoids the lengthy process of finding focus coordinate values and executing distance formulas explicitly.

Chapter Mix

Class 11 Mathematics: Conic Sections