Q1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Solution

Let the cross-section of the parabolic reflector be represented by a parabola whose vertex is at the origin and whose axis lies along the positive \(x\)-axis. The standard equation of such a parabola is \[ y^2 = 4ax, \] where \(a\) is the distance of the focus from the vertex.

The reflector has a diameter of \(20\text{ cm}\), so the radius of the opening is \(10\text{ cm}\). It is given that the depth of the reflector is \(5\text{ cm}\), meaning the rim of the reflector lies at the point where \(x = 5\) and \(y = \pm 10\).

Substituting \(x = 5\) and \(y = 10\) into the equation of the parabola, we get: \[ \begin{aligned} y^2 &= 4ax \\ 10^2 &= 4a(5) \\ 100 &= 20a \\ a &= 5 \end{aligned} \]

Hence, the focus of the parabolic reflector is at the point \((5,\,0)\), which means it lies \(5\text{ cm}\) from the vertex along the axis of the reflector.

Q2. 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

Let the vertex of the parabola be at the origin, and let the axis of the parabola be vertical. The standard equation of such a parabola is \[ x^2 = 4ay, \] where \(a\) is the distance of the focus from the vertex.

The arch is \(10\text{ m}\) high, so the base lies at \(y = 10\). The width at the base is \(5\text{ m}\), which means the points on the parabola at the base are \((\pm 2.5, 10)\).

Substituting \(x = 2.5\) and \(y = 10\) into the equation, we get: \[ \begin{aligned} x^2 &= 4ay \\ (2.5)^2 &= 4a(10) \\ 6.25 &= 40a \\ a &= \frac{6.25}{40} = \frac{5}{32} \end{aligned} \]

Now we find the width \(2\text{ m}\) from the vertex, that is at \(y = 2\). Substituting into the parabola equation: \[ \begin{aligned} x^2 &= 4ay \\ x^2 &= 4\left(\frac{5}{32}\right)(2) \\ x^2 &= \frac{40}{32} = \frac{5}{4} \\ x &= \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \end{aligned} \]

Since the width is the distance between the two symmetric points on either side of the axis, the total width is: \[ \text{Width} = 2x = 2 \cdot \frac{\sqrt{5}}{2} = \sqrt{5}\text{ m} \]

Therefore, the arch is \(\sqrt{5}\text{ m}\) wide at a height of \(2\text{ m}\) from the vertex.

Q3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Solution

Let the middle of the bridge roadway be taken as the origin, with the \(x\)-axis along the roadway and the \(y\)-axis representing the length of the vertical supporting wires. Since the cable hangs in the form of a parabola, the wire length varies according to a quadratic equation. The shortest wire is at the center and has length \(6\text{ m}\), so the vertex of the parabola is at \((0,6)\).

Hence, the equation of the parabola can be written as \[ y = ax^2 + 6. \] The roadway is \(100\text{ m}\) long, so the ends are at \(x = \pm 50\). The longest wire at each end has length \(30\text{ m}\), giving the point \((50,30)\) on the parabola.

Substituting \(x = 50\) and \(y = 30\), we determine \(a\): \[ \begin{aligned} 30 &= a(50)^2 + 6 \\ 30 &= 2500a + 6 \\ 24 &= 2500a \\ a &= \frac{24}{2500} \end{aligned} \]

Now, to find the length of the supporting wire \(18\text{ m}\) from the middle, we substitute \(x = 18\): \[ \begin{aligned} y &= a(18)^2 + 6 \\ y &= \frac{24}{2500} \cdot 324 + 6 \\ y &= 3.1104 + 6 \\ y &\approx 9.11 \end{aligned} \]

Therefore, the length of the supporting wire attached to the roadway \(18\text{ m}\) from the middle is approximately \(9.11\text{ m}\).

Q4. 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

Let the arch be represented by a semi-ellipse centered at the origin, with its major axis along the horizontal direction. Since the arch is \(8\text{ m}\) wide, the semi-major axis is \[ a = \frac{8}{2} = 4. \] The height at the center is \(2\text{ m}\), so the semi-minor axis is \[ b = 2. \] Thus, the equation of the ellipse is \[ \frac{x^2}{16} + \frac{y^2}{4} = 1, \] and we consider only the upper half since the arch is a semi-ellipse.

A point \(1.5\text{ m}\) from one end of the arch lies horizontally at a distance \[ x = 4 - 1.5 = 2.5 \] from the center.

Substituting \(x = 2.5\) into the ellipse equation, we get: \[ \begin{aligned} \frac{(2.5)^2}{16} + \frac{y^2}{4} &= 1 \\ \frac{6.25}{16} + \frac{y^2}{4} &= 1 \\ \frac{y^2}{4} &= 1 - \frac{6.25}{16} \\ \frac{y^2}{4} &= \frac{9.75}{16} \\ y^2 &= \frac{39}{16} \\ y &= \frac{\sqrt{39}}{4} \end{aligned} \]

Hence, the height of the arch at a point \(1.5\text{ m}\) from one end is \[ \frac{\sqrt{39}}{4}\text{ m} \approx 1.56\text{ m}. \]

Q5. 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

Let the rod touch the coordinate axes at points \(A(a,0)\) on the \(x\)-axis and \(B(0,b)\) on the \(y\)-axis. Since the rod has a fixed length of \(12\text{ cm}\), the distance between \(A\) and \(B\) gives \[ \begin{aligned} AB^2 &= a^2 + b^2 = 12^2 = 144. \end{aligned} \]

A point \(P\) lies on the rod such that it is \(3\text{ cm}\) from the end in contact with the \(x\)-axis. Thus, if \(AP = 3\) and \(PB = 9\), the point \(P\) divides the segment \(AB\) internally in the ratio \(3:9 = 1:3\).

Using the section formula, the coordinates of \(P\) are \[ \begin{aligned} P &= \left( \frac{3a + 1\cdot 0}{1+3}, \frac{3\cdot 0 + 1\cdot b}{1+3} \right) = \left( \frac{3a}{4}, \frac{b}{4} \right). \end{aligned} \] Let the coordinates of \(P\) be \((x,y)\). Then \[ \begin{aligned} x &= \frac{3a}{4} \Rightarrow a = \frac{4x}{3}, \\ y &= \frac{b}{4} \Rightarrow b = 4y. \end{aligned} \]

Substituting into the condition \(a^2 + b^2 = 144\), we obtain \[ \begin{aligned} \left(\frac{4x}{3}\right)^2 + (4y)^2 &= 144 \\ \frac{16x^2}{9} + 16y^2 &= 144 \\ \frac{x^2}{9} + y^2 &= 9. \end{aligned} \]

Hence, the locus of the point \(P\) is the ellipse given by \[ \frac{x^2}{9} + y^2 = 9. \]

Q6. 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

The given parabola is \[ x^2 = 12y, \] which is in the standard form \(x^2 = 4ay\). Comparing, we get \[ 4a = 12 \Rightarrow a = 3. \] Thus, the vertex of the parabola is at \((0,0)\).

The endpoints of the latus rectum of \(x^2 = 4ay\) are given by \[ (\pm 2a,\, a). \] Substituting \(a = 3\), the endpoints are \[ (6,3) \quad \text{and} \quad (-6,3). \]

The triangle is formed by joining the vertex \((0,0)\) to these two endpoints. The base of the triangle is the segment joining \((-6,3)\) and \((6,3)\), whose length is \[ \begin{aligned} \text{Base} &= 6 - (-6) = 12. \end{aligned} \]

The height of the triangle is the perpendicular distance from the vertex \((0,0)\) to the line \(y = 3\), which is \[ \text{Height} = 3. \]

Therefore, the area of the triangle is \[ \begin{aligned} \text{Area} &= \frac{1}{2} \times \text{Base} \times \text{Height} \\ &= \frac{1}{2} \times 12 \times 3 \\ &= 18. \end{aligned} \]

Hence, the required area of the triangle is \(18\) square units.

Q7. 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

Let the two flag posts be fixed points on a straight line, and let their midpoint be taken as the origin. The man observes that the sum of his distances from the two posts is always \(10\text{ m}\). This defines an ellipse, where the flag posts act as the foci.

The distance between the two flag posts is \(8\text{ m}\), so if the foci are at \((\pm c, 0)\), then \[ 2c = 8 \Rightarrow c = 4. \] The constant sum of distances from the foci is \(10\text{ m}\), which gives \[ 2a = 10 \Rightarrow a = 5, \] where \(a\) is the semi-major axis.

For an ellipse, the relationship between \(a\), \(b\), and \(c\) is \[ b^2 = a^2 - c^2. \] Substituting the known values, we obtain: \[ \begin{aligned} b^2 &= 5^2 - 4^2 \\ b^2 &= 25 - 16 \\ b^2 &= 9. \end{aligned} \]

Hence, the equation of the path traced by the man, which is the ellipse with center at the origin and major axis along the \(x\)-axis, is \[ \frac{x^2}{25} + \frac{y^2}{9} = 1. \]

Therefore, the man moves along the ellipse given by \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

Q8. An equilateral triangle is inscribed in the parabola \(y^2 = 4 ax\), where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution

The given parabola is \[ y^2 = 4ax, \] whose vertex is at the origin \((0,0)\). One vertex of the equilateral triangle lies at this point. Let the other two vertices be symmetric about the \(x\)-axis and lie on the parabola at \((x, y)\) and \((x, -y)\).

The side of the triangle between these two symmetric points has length \[ 2y. \] The distance from the vertex \((0,0)\) to the point \((x,y)\) is \[ \sqrt{x^2 + y^2}. \] Since the triangle is equilateral, all sides are equal, giving \[ \sqrt{x^2 + y^2} = 2y. \]

Squaring both sides, we obtain \[ \begin{aligned} x^2 + y^2 &= 4y^2 \\ x^2 &= 3y^2 \\ x &= \sqrt{3}\,y. \end{aligned} \]

Substituting \(x = \sqrt{3}\,y\) into the parabola equation \(y^2 = 4ax\), we get \[ \begin{aligned} y^2 &= 4a(\sqrt{3}\,y) \\ y &= 4a\sqrt{3}. \end{aligned} \]

Therefore, the side length of the equilateral triangle is \[ 2y = 2(4a\sqrt{3}) = 8a\sqrt{3}. \]

Hence, the length of each side of the equilateral triangle is \(8a\sqrt{3}\).