Equation of two diameters of a circle are 2x-3y=5$2x-3y=5$ and 3x-4y=7$3x-4y=7$. The line joining the points (-(22)/(7),-4)$(-\frac{22}{7},-4)$ and (-(1)/(7),3)$(-\frac{1}{7},3)$ intersects the circle at only one point P(α,β)$P(\alpha,\beta)$. Then 17β-α$17\beta-\alpha$ is equal to

Related Formula

Core Logic

We find the parametric parameters of coordinates P$P$ and Q$Q$, obtain their Cartesian values, and then apply the section formula with the focus S$S$ to calculate the splitting ratio.

Step 1: Find coordinates of P and Q

For parabola y² = 16x$y^2 = 16x$, the focal parameter is a = 4$a = 4$. Focus is S(4, 0)$S(4, 0)$. Let P$P$ be (a t₁², 2a t₁) = (1, -4)$(a t_1^2, 2a t_1) = (1, -4)$:

Since PQ$PQ$ is a focal chord, the parametric points are coupled:

Now, calculate the coordinates of Q$Q$:

Step 2: Solve for the dividing ratio

Let the focus S(4, 0)$S(4, 0)$ divide the line segment PQ$PQ$ internally in the ratio λ : 1$\lambda : 1$. Using the y$y$-coordinate of the section formula:

Thus, the focus S$S$ divides the chord internally in the ratio 1:4$1:4$. Since (1, 4) = 1$\gcd(1, 4) = 1$, we have m = 1$m = 1$ and n = 4$n = 4$:

Pattern Recognition

Harmonic Mean Shortcut: In any parabola, the focus divides a focal chord internally into segments of lengths SP$SP$ and SQ$SQ$ such that the semi-latus rectum 2a$2a$ is the harmonic mean of these segments: (1)/(SP) + (1)/(SQ) = (1)/(a)$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For a standard hyperbola: Focus: (± ae, 0)$(\pm ae, 0)$ Directrix: x = ± (a)/(e)$x = \pm \frac{a}{e}$ Eccentricity relation: (ae)² = a² + b²$(ae)^2 = a^2 + b^2$ Length of latus rectum: l = (2b²)/(a)$l = \frac{2b^2}{a}$

Core Logic

Given ae = √(10)$ae = \sqrt{10}$ and (a)/(e) = 9√(10)$\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these gives a²$a^2$, which determines both parameters.

Step 1: Find a and e

Substitute a = 3$a = 3$ into ae = √(10)$ae = \sqrt{10}$:

Step 2: Find b and l

Using (ae)² = a² + b²$(ae)^2 = a^2 + b^2$:

Then the length of latus rectum l$l$ is:

Step 3: Evaluate Final Expression

Calculate 9(e² + l)$9(e^2 + l)$:

Pattern Recognition

Multiplying focus location by directrix location immediately eliminates e$e$, giving a²$a^2$ directly. Once a²$a^2$ is known, b²$b^2$ follow seamlessly via (ae)² = a²+b²$(ae)^2 = a^2+b^2$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

Focal distances of any point P(a θ, b θ)$P(a\cos\theta, b\sin\theta)$ on an ellipse are given by:

Product of focal distances:

Core Logic

Compute the eccentricity e$e$, express the product SP · S'P$SP \cdot S'P$ in terms of ²θ$\cos^2\theta$, and analyze its bounds across the domain to find minimum and maximum limits.

Step 1: Determine Ellipse Parameters

Given a² = 18$a^2 = 18$ and b² = 9$b^2 = 9$.

Step 2: Express Focal Product

The parametric coordinates are P(3√(2) θ, 3 θ)$P(3\sqrt{2}\cos\theta, 3\sin\theta)$.

Since a²=18$a^2=18$ and (ae)² = a²-b² = 18-9 = 9$(ae)^2 = a^2-b^2 = 18-9 = 9$:

Step 3: Evaluate Extrema and Sum

Since 0 ≤ ²θ ≤ 1$0 \le \cos^2\theta \le 1$:

• Maximum value occurs when ²θ = 0 = 18$\cos^2\theta = 0 \implies \max = 18$.
• Minimum value occurs when ²θ = 1 = 18 - 9 = 9$\cos^2\theta = 1 \implies \min = 18 - 9 = 9$.

Pattern Recognition

The product of focal distances can also be written directly as b²$b^2$ at the minor axis vertices (max) and a²(1-e²)$a^2(1-e^2)$ varying down to a²-c²$a^2-c^2$. Summing them up yields b² + a² = 9 + 18 = 27$b^2 + a^2 = 9 + 18 = 27$ instantly.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For a standard parabola y² = 4ax$y^2 = 4ax$: Focus: S(a, 0)$S(a, 0)$ Parametric coordinates: (at², 2at)$(at^2, 2at)$ Equation of a circle on diametric endpoints (x₁, y₁)$(x_1, y_1)$ and (x₂, y₂)$(x_2, y_2)$:

Core Logic

Find the point P$P$ using the slope of the focal chord, write the equation of the circle with diameter PS$PS$, and find its y$y$-intercept.

Step 1: Determine P Coordinates

For y² = 4x$y^2 = 4x$, parameter a=1 S(1,0)$a=1 \implies S(1,0)$ and P(t², 2t)$P(t^2, 2t)$. Slope of focal chord PS$PS$:

Since P$P$ is in the first quadrant, t > 0 t = √(3)$t > 0 \implies t = \sqrt{3}$. Thus, P((√(3))², 2√(3)) = P(3, 2√(3))$P((\sqrt{3})^2, 2\sqrt{3}) = P(3, 2\sqrt{3})$.

Step 2: Construct the Diametric Circle Equation

Endpoints are S(1,0)$S(1,0)$ and P(3, 2√(3))$P(3, 2\sqrt{3})$:

Step 3: Solve for y-intercept

The circle touches/intersects the y$y$-axis at x = 0$x = 0$:

This is a perfect square expression (y - √(3))² = 0 y = √(3)$(y - \sqrt{3})^2 = 0 \implies y = \sqrt{3}$. Thus, the intercept value is a = √(3)$a = \sqrt{3}$.

Step 4: Compute Final Target Value

Pattern Recognition

A circle whose diameter is a focal radius always touches the tangent at the vertex (y$y$-axis for a standard parabola). The coordinate of the contact point is simply given by a t = 1 · √(3) = √(3)$a t = 1 \cdot \sqrt{3} = \sqrt{3}$, bypasses the full equation construction entirely.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

Perpendicular distance from point (x₀, y₀)$(x_0, y_0)$ to line Ax + By + C = 0$Ax + By + C = 0$:

Core Logic

Since the circle touches two symmetric intersecting lines, its center must lie on their angle bisector (x$x$-axis). Use this property to find the center parameters.

Step 1: Establish Center and Radius Equations

The lines are x+y-3=0$x+y-3=0$ and x-y-3=0$x-y-3=0$. The intersection point is (3,0)$(3,0)$, and the bisector line is the x$x$-axis. Let the center be C(a, 0)$C(a, 0)$. The radius r$r$ is the perpendicular distance to either line:

Step 2: Apply Point Passage Constraint

The circle equation is (x - a)² + y² = r²$(x - a)^2 + y^2 = r^2$. Substitute the given passage point (-9, 4)$(-9, 4)$:

Step 3: Solve for Quadratic Roots

Factor the quadratic equation:

Step 4: Compute Radii Squares Difference

Find the corresponding radius value for each root:

The absolute difference between their squares is:

Pattern Recognition

Recognizing that the center must lie on the line of symmetry (x$x$-axis) eliminates one variable parameter immediately, reducing a difficult geometric system to a simple single-variable quadratic equation.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines