Let r$r$ be the radius of the circle, which touches x$x$ -axis at point (a, 0)$(a, 0)$ , a < 0$a < 0$ and the parabola y² = 9x$y^2 = 9x$ at the point (4, 6)$(4, 6)$ . Then r$r$ is equal to

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

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