Learn Miscellaneous Exercise 11 with Free Lessons & Tips

The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the negative y-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form x² = −-4ay (as it is opening downwards).

It can be clearly seen that the parabola passes through point (52,−10)52,-10.

(52)2=−4a(−10)522=-4a-10

⇒4a=254×10=58⇒4a=254×10=58

Therefore, the arch is in the form of a parabola whose equation is x2=−58yx2=-58y.

When y = −-2, x2=−58(−2)x2=-58-2

⇒x2=54⇒x2=54

⇒x=5√2⇒x=52
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∴AB = 2×5√2m =5–√ m = 2.23 m (approx.)∴AB = 2×52m =5 m = 2.23 m (approx.)

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis. This can be diagrammatically represented as

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, 18 m from the middle.

Here, AB = 30 m, OC = 6 m, and.

The equation of the parabola is of the form x² = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 – 6) = (50, 24).

Since A (50, 24) is a point on the parabola,

∴Equation of the parabola,or 6x² = 625y

The x-coordinate of point D is 18.

Hence, at x = 18,

∴DE = 3.11 m

DF = DE + EF = 3.11 m + 6 m = 9.11 m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.

Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically represented as

The equation of the semi-ellipse will be of the form, where a is the semi-major axis

Accordingly, 2a = 8 ⇒ a = 4

b = 2

Therefore, the equation of the semi-ellipse is

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC⊥ OB.

OA = (4 – 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we obtain

∴AC = 1.56 m

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP = 3 cm.

Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm]

From P, draw PQ⊥OY and PR⊥OX.

In ΔPBQ,

In ΔPRA,

Thus, the equation of the locus of point P on the rod is.

The given parabola is x² = 12y.

On comparing this equation with x² = 4ay, we obtain 4a = 12 ⇒ a = 3

∴The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x² = 12 (3) ⇒ x² = 36 ⇒ x = ±6

∴The coordinates of A are (–6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of ΔOAB are O (0, 0), A (–6, 3), and B (6, 3).

Thus, the required area of the triangle is 18 unit².

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form, where a is the semi-major axis

Accordingly, 2a = 10 ⇒ a = 5

Distance between the foci (2c) = 8

⇒ c = 4

On using the relation, we obtain

Thus, the equation of the path traced by the man is.