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

The general equation of a circle is:

Radius of the circle:

If a set of points lies on this circle, their coordinates must satisfy the equation.

Core Logic

Since (0,0)$(0,0)$ lies on the circle:

Thus, the equation simplifies to:

Step 1: Finding g$g$, f$f$ and r²$r^2$

Substitute (4,6)$(4,6)$:

Substitute (-1,5)$(-1,5)$:

Substituting g$g$ from (2) into (1):

The circle equation is:

Calculating radius squared r²$r^2$:

Step 2: Solving for k$k$

The point (k, 3k)$(k, 3k)$ lies on this circle:

Since the points must be distinct and k=0$k=0$ gives (0,0)$(0,0)$ which is already a given point, we must have:

Now, calculate 10k + r²$10k + r^2$:

Pattern Recognition

Notice that the slope of the line joining origin (0,0)$(0,0)$ to the general point is y = 3x$y = 3x$. For three given coordinates, if origin is one of them, the circle equation lacks the constant c$c$. It is always faster to first solve for parameters g, f$g, f$ and then check geometry.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 10 Mathematics: Coordinate Geometry

Related Formula

For an ellipse E$E$:

• Foci: (± ae, 0)$(\pm ae, 0)$
• Eccentricity: b² = a²(1 - e²)$b^2 = a^2(1 - e^2)$
• The circle of minimum area enclosing a centered ellipse has diameter equal to the major axis of the ellipse (R = a$R = a$).
• Area of triangle: Area = (1)/(2) · base · height$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

Core Logic

Let's first find coordinates a$a$ and b$b$:

• ae = 2$ae = 2$
• e = (1)/(2) a((1)/(2)) = 2 a = 4$e = \frac{1}{2} \implies a\left(\frac{1}{2}\right) = 2 \implies a = 4$
• b² = a²(1 - e²) = 16(1 - (1)/(4)) = 12 b = 2√(3)$b^2 = a^2(1 - e^2) = 16\left(1 - \frac{1}{4}\right) = 12 \implies b = 2\sqrt{3}$

Step 1: Setting Circle and Triangle geometry

The enclosing circle C$C$ has radius R = a = 4$R = a = 4$, centered at (0,0)$(0,0)$. Thus, its equation is:

The intersection of the ellipse with the negative y$y$-axis is (0, -b) = (0, -2√(3))$(0, -b) = (0, -2\sqrt{3})$.

Since side QR$QR$ (length = 8$= 8$) is parallel to the major axis (x$x$-axis) and contains (0, -2√(3))$(0, -2\sqrt{3})$, the equation of the line containing QR$QR$ is: y = -2√(3)$y = -2\sqrt{3}$

Step 2: Maximizing Area of Δ PQR$\Delta PQR$

The perpendicular height of vertex P(4 θ, 4 θ)$P(4\cos\theta, 4\sin\theta)$ from the base line y = -2√(3)$y = -2\sqrt{3}$ is:

To maximize the area, we maximize height H$H$ by choosing θ = 1$\sin\theta = 1$:

Pattern Recognition

The enclosing circle with minimum area is called the auxiliary circle. Its radius is equal to the semi-major axis a$a$. Max height of a triangle with a base fixed at line y=-k$y=-k$ and vertex on the circle is R + k$R + k$. This directly gives Area = (1)/(2) · base · (a+b)$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot (a+b)$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Related Formula

For a circle x² + (y-k)² = R²$x^2 + (y-k)^2 = R^2$ and any smooth curve, the shortest distance lies along the normal to the curve passing through the center of the circle C(h,k)$C(h,k)$:

where P$P$ is the point of normal intersection on the curve.

Core Logic

Let's first identify the circle parameters:

Center C = (0, -6)$C = (0, -6)$ and radius R = 1$R = 1$.

The first curve is the parabola y² = 8x$y^2 = 8x$, where a = 2$a = 2$.

Normal equation of y² = 4ax$y^2 = 4ax$ in slope form:

Substituting a=2$a=2$:

Step 1: Find normal passing through circle center

Normal passes through C(0, -6)$C(0, -6)$:

By inspection, m=1$m=1$ is a real solution:

Since m² + m + 3 = 0$m^2 + m + 3 = 0$ has complex roots, the unique real normal slope is m=1$m=1$.

Step 2: Point calculation and shortest distance

For m=1$m=1$ and a=2$a=2$, normal intersection point P(am², -2am)$P(am^2, -2am)$ is:

Distance from P(2,-4)$P(2,-4)$ to center C(0,-6)$C(0,-6)$:

Shortest distance:

Pattern Recognition

The shortest distance between a parabola and a circle is always along the common normal of the parabola passing through the circle's center. Finding the normal in slope form and solving for m$m$ avoids complex calculus.

Chapter Mix

Class 11 Mathematics: Conic Sections

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