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Conic Sections - Sections of a Cone

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A conic section is a curve obtained as the intersection of a plane with a double-napped right circular cone. Imagine two identical cones joined at their vertices (the apex), extending infinitely in opposite directions along a common vertical axis. The angle between the axis and the generator (the line that rotates to form the cone) is called the semi-vertical angle, denoted by α\alpha.

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The shape of the conic section depends on the angle β\beta made by the intersecting plane with the axis of the cone. Visualise a flat sheet (the plane) slicing through the cone at different tilts; as the tilt changes, the boundary of the slice changes from a circle to an ellipse, a parabola, or a hyperbola.

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A Circle is formed when the plane is perpendicular to the axis, meaning β=90∘\beta = 90^{\circ}. Visually, this is a perfectly horizontal cut through one of the naps, resulting in a round shape where every point on the edge is equidistant from the axis.

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An Ellipse is formed when the cutting plane is slightly tilted such that the angle β\beta satisfies α<β<90∘\alpha < \beta < 90^{\circ}. Visually, the plane cuts through all generators of a single nap but at an angle, resulting in an elongated, closed oval shape.

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A Parabola is formed when the angle β\beta is exactly equal to the semi-vertical angle α\alpha (β=α\beta = \alpha). In this case, the plane is parallel to one of the generator lines. Visually, this creates an open-ended U-shaped curve that extends infinitely in one direction within one nap of the cone.

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A Hyperbola is formed when the plane cuts through both the upper and lower naps of the cone. This occurs when 0≤β<α0 \le \beta < \alpha, meaning the plane is parallel to the axis or tilted very steeply. Visually, this results in two separate, unbounded curves that open in opposite directions.

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Degenerate Conics occur when the plane passes through the vertex of the cone. Depending on the angle β\beta, the section may degenerate into a single point (if β>α\beta > \alpha), a single straight line (if β=α\beta = \alpha), or two intersecting straight lines (if 0≤β<α0 \le \beta < \alpha).

📐Formulae

Standard equation of a circle with center (h,k)(h, k) and radius rr: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Standard equation of a parabola (opening right): y2=4axy^2 = 4ax

Standard equation of a horizontal ellipse: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, where a>ba > b

Standard equation of a horizontal hyperbola: x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

Relationship between semi-major axis aa, semi-minor axis bb, and distance to focus cc for an ellipse: c2=a2−b2c^2 = a^2 - b^2

Relationship between transverse axis aa, conjugate axis bb, and distance to focus cc for a hyperbola: c2=a2+b2c^2 = a^2 + b^2

Eccentricity: e=cae = \frac{c}{a}

💡Examples

Problem 1:

Determine the type of conic section formed when a plane intersects a double-napped right circular cone with a semi-vertical angle of 35∘35^{\circ}, given that the angle between the plane and the axis is 50∘50^{\circ}.

Solution:

Step 1: Identify the given angles. Semi-vertical angle α=35∘\alpha = 35^{\circ}. Angle of the plane with the axis β=50∘\beta = 50^{\circ}. Step 2: Compare α\alpha and β\beta. Here, 35∘<50∘<90∘35^{\circ} < 50^{\circ} < 90^{\circ}, which means α<β<90∘\alpha < \beta < 90^{\circ}. Step 3: Based on the geometric definition, when α<β<90∘\alpha < \beta < 90^{\circ}, the plane cuts through only one nap and is not perpendicular to the axis.

Explanation:

Since the angle β\beta is greater than α\alpha but less than 90∘90^{\circ}, the resulting conic section is an ellipse.

Problem 2:

Find the equation of a circle that represents a section of a cone, given that its center is at (0,0)(0, 0) and it passes through the point (3,4)(3, 4).

Solution:

Step 1: Use the standard form of the circle equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. Since the center is (0,0)(0,0), h=0h=0 and k=0k=0. Step 2: The equation simplifies to x2+y2=r2x^2 + y^2 = r^2. Step 3: Substitute the point (3,4)(3, 4) into the equation to find r2r^2. 32+42=r2  ⟹  9+16=r2  ⟹  r2=253^2 + 4^2 = r^2 \implies 9 + 16 = r^2 \implies r^2 = 25. Step 4: Write the final equation: x2+y2=25x^2 + y^2 = 25.

Explanation:

This problem applies the distance formula (or circle equation) to find the radius from a given point and center, then substitutes it back into the standard form.