Rbse Solutions for Class 11 maths Chapter 10 Miscellaneous Exercise | Conic Sections

Get detailed, step-by-step solutions for the NCERT Class 11 Maths Chapter 10 Miscellaneous Exercise . Solve application problems for parabolas including finding the focus of a reflector (Q.1), and determining the width/height of parabolic arches (Q.2, 3). Find the height of a semi-elliptical arch (Q.4) and the locus of a moving rod (Q.5). Analyze the elliptical path of a running man (Q.7) and properties of a triangle in a parabola (Q.6, 8).

1. Parabolic Reflector

Problem: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Solution:

1. Set up the equation: Place the vertex of the parabola at the origin $(0, 0)$ and its axis along the $x$-axis. Since the reflector opens to the side, its equation is $y^2 = 4ax$.
2. Use the dimensions: The diameter is 20 cm, so the distance from the axis to the rim is $20/2 = 10$ cm. The depth is $5$ cm. This means the parabola passes through the point $(5, 10)$.
3. Solve for $4a$: Substitute the point $(5, 10)$ into the equation:$$10^2 = 4a(5)$$$$100 = 20a$$$$a = \frac{100}{20} = 5$$
4. Find the focus: The focus is at $(a, 0)$.
   • Focus: $\mathbf{(5, 0)}$

The focus of the parabolic reflector is 5 cm from the vertex.

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2. Parabolic Arch

Problem: 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?

Solution:

1. Set up the equation: Place the vertex at the origin $(0, 0)$ and the axis along the $y$-axis. Since the arch opens downward (it must be supported at the base), the equation is $x^2 = -4ay$.
2. Find the endpoints: The arch is 5 m wide, so the base extends from $x = -2.5$ to $x = 2.5$. The height is 10 m, so the endpoints of the arch are $(-2.5, -10)$ and $(2.5, -10)$. Use the point $(2.5, -10)$.
3. Solve for $4a$: Substitute the point $(2.5, -10)$ into the equation:$$(2.5)^2 = -4a(-10)$$$$6.25 = 40a$$$$4a = \frac{6.25}{10} = 0.625$$The equation is $x^2 = -0.625y$.
4. Find the width 2 m from the vertex: The point 2 m from the vertex corresponds to $y = -2$. Let the width at this height be $2x_1$.$$x_1^2 = -0.625(-2)$$$$x_1^2 = 1.25$$$$x_1 = \sqrt{1.25} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \text{ m}$$
5. Calculate the width: The total width is $2x_1$.$$\text{Width} = 2 \cdot \frac{\sqrt{5}}{2} = \sqrt{5} \text{ m}$$Using $\sqrt{5} \approx 2.236$:The arch is $\sqrt{5}$ meters (approximately 2.236 m) wide at 2 m from the vertex.

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3. Suspension Bridge Cable

Problem: The cable of a suspension bridge hangs in the form of a parabola. The horizontal roadway is 100 m long. The longest vertical wire is 30 m and the shortest is 6 m. Find the length of a supporting wire attached 18 m from the middle.

Solution:

1. Set up the equation: Place the vertex (where the shortest wire is) at $(0, 6)$. The axis is the $y$-axis. Since the cable opens upward, the equation is $x^2 = 4a(y – 6)$.
2. Use the endpoints: The roadway is 100 m long, so the endpoints of the cable are $x = \pm 50$. The longest wire is 30 m, so the cable passes through the points $(-50, 30)$ and $(50, 30)$. Use $(50, 30)$.
3. Solve for $4a$: Substitute $(50, 30)$ into the equation:$$50^2 = 4a(30 – 6)$$$$2500 = 4a(24)$$$$4a = \frac{2500}{24} = \frac{625}{6}$$The equation is $x^2 = \frac{625}{6}(y – 6)$.
4. Find the wire length at $x = 18$: We need to find the $y$-coordinate when $x = 18$.$$18^2 = \frac{625}{6}(y – 6)$$$$324 = \frac{625}{6}(y – 6)$$$$y – 6 = 324 \cdot \frac{6}{625} = \frac{1944}{625} \approx 3.1104$$$$y = 6 + \frac{1944}{625} = \frac{3750 + 1944}{625} = \frac{5694}{625}$$
5. Final length: The length of the supporting wire is $y$.
   • Length: $\frac{5694}{625}$ meters (approximately 9.1104 m).

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4. Semi-Elliptical Arch

Problem: An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Solution:

1. Set up the equation: Place the center of the ellipse at the origin $(0, 0)$. Since the arch is a semi-ellipse, we only consider the upper half (where $y \ge 0$). The equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
2. Determine $a$ and $b$:
   • Width ($2a$): $2a = 8 \text{ m} \implies a = 4 \text{ m} \implies a^2 = 16$.
   • Height ($b$): $b = 2 \text{ m} \implies b^2 = 4$.The equation is $\frac{x^2}{16} + \frac{y^2}{4} = 1$.
3. Find the $x$-coordinate: The question asks for the height at a point 1.5 m from one end. The ends of the arch are at $x = \pm 4$.
   • If we measure from the end $x = 4$, the point is $x = 4 – 1.5 = 2.5$.
   • If we measure from the end $x = -4$, the point is $x = -4 + 1.5 = -2.5$.Due to symmetry, we use $x = 2.5$ m.
4. Solve for $y$ (the height): Substitute $x = 2.5$ into the equation:$$\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1$$$$\frac{6.25}{16} + \frac{y^2}{4} = 1$$$$\frac{y^2}{4} = 1 – \frac{6.25}{16} = \frac{16 – 6.25}{16} = \frac{9.75}{16}$$$$y^2 = 4 \cdot \frac{9.75}{16} = \frac{9.75}{4} = \frac{39/4}{4} = \frac{39}{16}$$$$y = \sqrt{\frac{39}{16}} = \frac{\sqrt{39}}{4} \text{ m}$$
5. Final height:
   • Height: $\frac{\sqrt{39}}{4}$ meters (approximately 1.56 m).

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5. Locus of a Point on a Moving Rod

Problem: A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is 3 cm from the end in contact with the $x$-axis.

Solution:

1. Set up coordinates: Let the rod have length $L = 12$. Let the ends touching the axes be $A(a, 0)$ and $B(0, b)$. The rod length equation is $a^2 + b^2 = L^2 = 144$.
2. Define point $P$: Let $P(x, y)$ be the point on the rod. The point $P$ is 3 cm from $A$ and $12 – 3 = 9$ cm from $B$.
3. Use Section Formula (or similar triangles): Point $P$ divides the rod $AB$ in the ratio $AP : PB = 3 : 9 = 1 : 3$.Using the section formula for $P(x, y)$ dividing $AB$ in the ratio $1 : 3$:$$x = \frac{3(a) + 1(0)}{3 + 1} = \frac{3a}{4} \implies \mathbf{a = \frac{4x}{3}}$$$$y = \frac{3(0) + 1(b)}{3 + 1} = \frac{b}{4} \implies \mathbf{b = 4y}$$
4. Find the locus: Substitute the expressions for $a$ and $b$ into the rod length equation $a^2 + b^2 = 144$:$$\left(\frac{4x}{3}\right)^2 + (4y)^2 = 144$$$$\frac{16x^2}{9} + 16y^2 = 144$$Divide the entire equation by 16:$$\frac{x^2}{9} + y^2 = 9$$Finally, divide by 9 to get the standard form:$$\mathbf{\frac{x^2}{81} + \frac{y^2}{9} = 1}$$The locus of point $P$ is an ellipse with $a=9$ and $b=3$.

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6. Area of Triangle in a Parabola

Problem: Find the area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum.

Solution:

1. Find the parameters: The parabola is $x^2 = 4ay$. Comparing with $x^2 = 12y$:$$4a = 12 \implies \mathbf{a = 3}$$
2. Find the coordinates:
   • Vertex ($V$): $(0, 0)$.
   • Focus ($F$): $(0, a) = (0, 3)$.
   • Ends of Latus Rectum ($L, R$): The latus rectum is perpendicular to the $y$-axis and passes through the focus $(0, 3)$. The length is $4a = 12$. The endpoints are at a distance $2a = 6$ from the focus along the $x$-axis.$$L = (-6, 3) \quad \text{and} \quad R = (6, 3)$$
3. Find the area of $\triangle VLR$: The triangle has vertices $V(0, 0)$, $L(-6, 3)$, and $R(6, 3)$.
   • Base: The base $LR$ is horizontal, its length is $6 – (-6) = 12$.
   • Height: The height is the perpendicular distance from $V$ to the line $y = 3$, which is $3 – 0 = 3$.$$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 12 \times 3$$
   • Area: 18 square units.

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7. Locus Traced by a Running Man

Problem: A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Solution:

1. Identify the conic section: The locus of a point (the man) such that the sum of its distances from two fixed points (flag posts/foci) is constant is the definition of an Ellipse

.

2. Determine parameters:

* Sum of distances ($2a$): $2a = 10 \text{ m} \implies \mathbf{a = 5}$.

* Distance between foci ($2c$): $2c = 8 \text{ m} \implies \mathbf{c = 4}$.

3. Find $b^2$: Use the ellipse relation $a^2 = b^2 + c^2$:

$$5^2 = b^2 + 4^2$$

$$25 = b^2 + 16$$

$$b^2 = 25 – 16 = \mathbf{9}$$

4. Write the equation: Assume the ellipse is centered at the origin and the foci are on the $x$-axis. The form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

$$\mathbf{\frac{x^2}{25} + \frac{y^2}{9} = 1}$$

The equation of the path traced by the man is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

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8. Equilateral Triangle Inscribed in a Parabola

Problem: An equilateral triangle is inscribed in the parabola $y^2 = 4ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution:

1. Set up coordinates: The parabola is $y^2 = 4ax$. Its vertex is $V(0, 0)$. Let the equilateral triangle be $VAB$.
2. Use symmetry: Since the parabola is symmetric about the $x$-axis, and one vertex is at $(0, 0)$, the side $AB$ must be perpendicular to the axis of the parabola ($x$-axis) and bisected by it.
3. Define vertices:
   • $V = (0, 0)$.
   • Let the length of the side be $s$.
   • Let $A = (x_1, y_1)$ and $B = (x_1, -y_1)$. The length of $AB$ is $2y_1$. Thus, $s = 2y_1$.
   • Since $A$ is on the parabola, $y_1^2 = 4ax_1$.
4. Use the equilateral property: The distance $VA$ must equal the side length $s$.$$VA^2 = s^2$$$$x_1^2 + y_1^2 = (2y_1)^2$$$$x_1^2 + y_1^2 = 4y_1^2$$$$x_1^2 = 3y_1^2 \implies \mathbf{x_1 = \sqrt{3}y_1}$$ (Since $x_1 > 0$)
5. Solve for $y_1$: Substitute $x_1$ into the parabola equation $y_1^2 = 4ax_1$:$$y_1^2 = 4a(\sqrt{3}y_1)$$Since $y_1 \ne 0$, we can divide by $y_1$:$$y_1 = 4a\sqrt{3}$$
6. Find the side length $s$:$$s = 2y_1 = 2(4a\sqrt{3}) = 8a\sqrt{3}$$The length of the side of the equilateral triangle is $8a\sqrt{3}$.

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