Class 11 Maths NCERT Solutions Chapter 10 Miscellaneous Exercise – Conic Sections

Placing the vertex of the reflector at the origin with its axis along the positive x-axis, the equation has the form y^2=4ax.

The diameter is 20 cm, so the rim is 10 cm above and below the axis; the depth is 5 cm, so the rim lies at x=5. The point (5,10) lies on the parabola:

10^2=4a(5) \ \Rightarrow\ 100=20a \ \Rightarrow\ a=5

Placing the vertex at the top of the arch and the axis vertical, the arch opens downward, so the equation has the form x^2=-4ay.

The base is 10 m below the vertex (y=-10) and 5 m wide there, so the base corners are at x=\pm2.5. Substituting (2.5,-10):

(2.5)^2=-4a(-10) \ \Rightarrow\ 6.25=40a \ \Rightarrow\ a=\dfrac{5}{32}

At 2 m from the vertex, y=-2:

x^2=4a(2)=4\left(\dfrac{5}{32}\right)(2)=\dfrac{5}{4}

x=\sqrt{\dfrac54}=\dfrac{\sqrt5}{2}, so the full width is 2x=\sqrt5.

The cable is lowest at the middle of the span (where the wire is shortest, 6 m) and rises symmetrically toward the two ends (where the wire is longest, 30 m). Placing the vertex of the parabola at the lowest point of the cable, with the axis vertical, the equation has the form x^2=4ay, where y measures the cable's height above its lowest point.

At the ends of the roadway, x=50 (half of the 100 m span), and the cable has risen by 30-6=24 m above its lowest point:

50^2=4a(24) \ \Rightarrow\ 2500=96a \ \Rightarrow\ a=\dfrac{625}{24}

At 18 m from the middle, x=18:

18^2=4\left(\dfrac{625}{24}\right)y \ \Rightarrow\ 324=\dfrac{2500}{24}y \ \Rightarrow\ y=\dfrac{324\times24}{2500}=\dfrac{1944}{625}

The wire length is the shortest wire (6 m) plus this rise:

6+\dfrac{1944}{625}=\dfrac{3750+1944}{625}=\dfrac{5694}{625}

Placing the centre of the base at the origin, with the arch as the upper half of an ellipse: the width 8 m gives a=4 (semi-major axis along the base), and the height 2 m at the centre gives b=2. The equation is:

\dfrac{x^2}{16}+\dfrac{y^2}{4}=1

A point 1.5 m from one end (say, the end at x=-4) is at x=-4+1.5=-2.5, i.e. x^2=6.25=\dfrac{25}{4}.

Substituting:

\dfrac{25/4}{16}+\dfrac{y^2}{4}=1 \ \Rightarrow\ \dfrac{25}{64}+\dfrac{y^2}{4}=1 \ \Rightarrow\ \dfrac{y^2}{4}=\dfrac{39}{64}

y^2=\dfrac{39}{16} \ \Rightarrow\ y=\dfrac{\sqrt{39}}{4}

Let the rod be AB, with A on the x-axis and B on the y-axis, and let \theta be the angle the rod makes with the x-axis. Let P be the point on the rod with AP=3, so PB=12-3=9.

Drawing PQ perpendicular to the y-axis and PR perpendicular to the x-axis, from the right triangles formed:

\cos\theta=\dfrac{x}{PB}=\dfrac{x}{9} \qquad \sin\theta=\dfrac{y}{AP}=\dfrac{y}{3}

Using \cos^2\theta+\sin^2\theta=1:

\left(\dfrac{x}{9}\right)^2+\left(\dfrac{y}{3}\right)^2=1

Comparing x^2=12y with x^2=4ay: 4a=12 \ \Rightarrow\ a=3. The focus is at (0,3), and the ends of the latus rectum (the horizontal chord through the focus) are at y=3, x=\pm2a=\pm6, giving the points (6,3) and (-6,3).

The vertex is at (0,0). The triangle has base 6-(-6)=12 (along y=3) and height 3 (the perpendicular distance from the vertex to that line):

\text{Area}=\dfrac12\times12\times3=18

Since the sum of the distances from the man to the two fixed flag posts is always constant, the path traced is an ellipse with the flag posts as foci.

Placing the centre at the midpoint of the two flag posts, with the foci on the x-axis: the constant sum gives 2a=10 \ \Rightarrow\ a=5, and the distance between the posts gives 2c=8 \ \Rightarrow\ c=4.

b^2=a^2-c^2=25-16=9

By the symmetry of the parabola about the x-axis, if one vertex of the equilateral triangle is at the origin, the other two vertices must be reflections of each other across the x-axis: say (x_1,y_1) and (x_1,-y_1), both lying on the parabola, so y_1^2=4ax_1.

The side between these two points has length 2y_1. The side from the origin to (x_1,y_1) has length \sqrt{x_1^2+y_1^2}. For an equilateral triangle, these are equal:

\sqrt{x_1^2+y_1^2}=2y_1 \ \Rightarrow\ x_1^2+y_1^2=4y_1^2 \ \Rightarrow\ x_1^2=3y_1^2 \ \Rightarrow\ x_1=\sqrt3\,y_1

Substituting into y_1^2=4ax_1:

y_1^2=4a\sqrt3\,y_1 \ \Rightarrow\ y_1=4\sqrt3\,a (dividing by y_1\ne0)

The side length is:

2y_1=8\sqrt3\,a