Conic Sections Miscellaneous Ex: Class 11 NCERT Solutions & Concepts

Strategy for Solving Miscellaneous Problems

The miscellaneous exercise questions often seem tricky because they don't explicitly state the type of conic or provide information in the standard format. The key is to have a systematic approach. First, carefully read the problem and extract all given information: points, foci, vertices, length of an axis, directrix, etc. Your next crucial step is to identify the conic. Is the problem about a point moving equidistant from a fixed point and a line (parabola)? Or is the sum (ellipse) or difference (hyperbola) of its distances from two fixed points constant? Once you've identified the conic and its orientation (e.g., a horizontal ellipse or a vertical parabola), recall its standard equation. Use the given information to create equations and solve for the unknown parameters like 'a', 'b', or 'p'. Finally, substitute these values back into the standard equation to get your answer. This step-by-step process turns a complex problem into a manageable one.

Worked Examples from NCERT Miscellaneous Exercise

• Example 1 (Parabolic Reflector): A beam of light is originating from the focus of a parabolic reflector which is 20 cm in diameter and 5 cm deep. Find the equation of the parabola. Solution: Step 1: Visualize and set up the coordinate system. Let's place the vertex of the parabolic reflector at the origin (0, 0) and its axis along the x-axis. Since it opens to the right, the equation of the parabola is of the form y² = 4ax, where a > 0. Step 2: Use the given dimensions to find a point on the parabola. The reflector is 20 cm in diameter and 5 cm deep. This means the parabola passes through a point (x, y) where the depth is x = 5 cm. The diameter is 20 cm, so the distance from the axis to the edge is y = 20/2 = 10 cm. Thus, the point (5, 10) lies on the parabola. Step 3: Substitute the point into the standard equation to find 'a'. Substitute x = 5 and y = 10 into y² = 4ax: 10² = 4 a 5 100 = 20a a = 100 / 20 = 5 Step 4: Write the final equation. The equation of the parabola is y² = 4(5)x, which is y² = 20x. The focus is at (a, 0), which is (5, 0). So the light source is 5 cm from the vertex.
• Example 2 (Ellipse from definition): Find the equation for the ellipse that satisfies the conditions: Ends of major axis (0, ±√5), ends of minor axis (±1, 0). Solution: Step 1: Identify the orientation and center of the ellipse. The ends of the major axis are on the y-axis (0, ±√5), and the ends of the minor axis are on the x-axis (±1, 0). This means the ellipse is vertical. The center of the ellipse is the midpoint of the major axis, which is (0,0). Step 2: Determine the values of 'a' and 'b'. For a vertical ellipse, the vertices of the major axis are (0, ±a) and the vertices of the minor axis are (±b, 0). From the given information: Ends of major axis are (0, ±√5), so a = √5. Ends of minor axis are (±1, 0), so b = 1. Step 3: Write the standard equation for a vertical ellipse. The standard equation for a vertical ellipse centered at the origin is: x²/b² + y²/a² = 1 Step 4: Substitute the values of 'a' and 'b' into the equation. Substituting a = √5 and b = 1: x²/1² + y²/(√5)² = 1 x² + y²/5 = 1 Final Answer: The equation of the ellipse is x² + y²/5 = 1.

Exam Tip: Identifying the Conic is Half the Battle

In the exam, miscellaneous questions rarely name the conic. You must deduce it from the geometric properties provided. Memorize these key identifiers:

• Parabola: Look for the mention of a single focus and a single directrix. The fundamental property is that any point on the parabola is equidistant from these two.
• Ellipse: Look for two foci. The key property is that the sum of the distances from any point on the ellipse to the two foci is a constant (equal to the length of the major axis, 2a).
• Hyperbola: Look for two foci. The key property is that the difference of the distances from any point on the hyperbola to the two foci is a constant (equal to the length of the transverse axis, 2a).
• Eccentricity (e): If given, it's a dead giveaway: e = 1 for a parabola, 0 < e < 1 for an ellipse, and e > 1 for a hyperbola.

Practice Questions with Solutions

• Q: An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? A: Step 1: Set up the coordinate system. Let the vertex be at the origin (0,0) and the axis be along the y-axis. Since it's an arch opening downwards, the equation is of the form x² = -4ay. Step 2: Use the given dimensions to find a point on the parabola. The arch is 10m high and 5m wide. The endpoint of the base is at a y-coordinate of -10. The width is 5m, so the x-coordinate is 5/2 = 2.5. Thus, the point (2.5, -10) lies on the parabola. Step 3: Find the value of 'a'. Substitute (2.5, -10) into x² = -4ay. (2.5)² = -4a(-10) => 6.25 = 40a => a = 6.25/40 = 5/32. The equation is x² = -4(5/32)y = -(5/8)y. Step 4: Find the width at 2m from the vertex. We need to find the width when the height from the vertex is 2m, which means y = -2. Substitute y = -2 into the equation: x² = -(5/8)(-2) = 5/4. So, x = √(5/4) = √5/2. This is half the width. The full width is 2 * (√5/2) = √5 m. Final answer: The arch is √5 m wide at 2 m from the vertex.
• Q: Find the equation of the hyperbola where foci are (±5, 0) and the transverse axis is of length 8. A: Step 1: Identify the type and orientation of the hyperbola. The foci (±c, 0) are on the x-axis, so it is a horizontal hyperbola with its center at the origin (0,0). The standard equation is x²/a² - y²/b² = 1. Step 2: Determine the values of 'a' and 'c'. From the foci (±5, 0), we have c = 5. The length of the transverse axis is 2a = 8, so a = 4. Step 3: Find the value of 'b' using the relation c² = a² + b². We have 5² = 4² + b² => 25 = 16 + b² => b² = 25 - 16 = 9. So, b = 3. Step 4: Substitute 'a' and 'b' into the standard equation. x²/4² - y²/3² = 1. Final answer: The equation of the hyperbola is x²/16 - y²/9 = 1.
• Q: Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16. A: Step 1: Let the centre of the circle be (h, k). Since the centre lies on the line 4x + y = 16, we have 4h + k = 16. This is our first equation: k = 16 - 4h. Step 2: Use the property that the distance from the centre to any point on the circle is the radius (r). So, the distance from (h, k) to (4, 1) is equal to the distance from (h, k) to (6, 5). (h-4)² + (k-1)² = (h-6)² + (k-5)² Step 3: Expand and simplify the equation. h² - 8h + 16 + k² - 2k + 1 = h² - 12h + 36 + k² - 10k + 25. -8h - 2k + 17 = -12h - 10k + 61. 4h + 8k = 44 => h + 2k = 11. Step 4: Solve the two linear equations for h and k. We have h + 2k = 11 and k = 16 - 4h. Substitute k in the first equation: h + 2(16 - 4h) = 11 => h + 32 - 8h = 11 => -7h = -21 => h = 3. Now find k: k = 16 - 4(3) = 16 - 12 = 4. The centre is (3, 4). Step 5: Find the radius r². r² = (h-4)² + (k-1)² = (3-4)² + (4-1)² = (-1)² + 3² = 1 + 9 = 10. The equation of the circle is (x-h)² + (y-k)² = r². Final answer: The equation of the circle is (x-3)² + (y-4)² = 10.
• Q: Find the coordinates of the foci and the eccentricity of the ellipse given by the equation 25x² + 9y² = 225. A: Step 1: Convert the equation to standard form. Divide the entire equation by 225 to make the right side equal to 1. (25x²/225) + (9y²/225) = 1 => x²/9 + y²/25 = 1. Step 2: Identify the type of ellipse and the values of a² and b². The denominator of y² (25) is larger than the denominator of x² (9). So, it is a vertical ellipse. We have a² = 25 and b² = 9. This means a = 5 and b = 3. Step 3: Find the value of c using the relation c² = a² - b² for an ellipse. c² = 25 - 9 = 16 => c = 4. Step 4: Determine the coordinates of the foci. For a vertical ellipse, the foci are located at (0, ±c). Therefore, the foci are at (0, 4) and (0, -4). Step 5: Calculate the eccentricity 'e'. The formula for eccentricity is e = c/a. e = 4/5. Final answer: The foci are at (0, ±4) and the eccentricity is e = 4/5.

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