11类NCERT解决方案-第11章圆锥曲线部分-第11章的其他练习

问题1.如果抛物面反射镜的直径为20厘米，深度为5厘米，请找到焦点。

解决方案：

The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive axis,

The equation for this parabola will be, 

y² = 4ax, which passes through A(5,10)

So, 10² = 4 a(5)

a = $\frac{100}{20}$

a = 5

Hence, the focus of the parabola is (a, 0) = (5, 0), which is the mid – point of the diameter.

问题2：拱形为抛物线形，其轴垂直。拱门高10 m，底部宽5 m。离抛物线顶点2 m的宽度是多少?

解决方案：

We can take the origin of the coordinate plane as vertex of the arch, where its vertical axis is along the positive y – axis

The equation for this parabola will be,

x² = 4ay, which passes through A($\frac{5}{2}$,10)

So, $(\frac{5}{2})^2$ = 4 a(10)

4a = $\frac{25}{4*10}$

4a = $\frac{5}{8}$

Equation of the parabola becomes,

x² = $\frac{5}{8}$ y

We need to find width, when height = 2m.

y = 2, so

x² = $\frac{5}{8}$ × (2)

x² = $\frac{5}{4}$

x = $\frac{√5}{2}$

Width = AB

= 2 × $\frac{√5}{2}$

Width (AB) = √5 m

问题3.均匀负载的悬索桥的电缆以抛物线的形式悬挂。水平且长度为100 m的巷道由连接到电缆的垂直电线支撑，最长的电线为30 m，最短的电线为6 m。找到距离中间18 m的与巷道相连的支撑线的长度。

解决方案：

According to the given conditions, we can diagrammatically represent

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

So, AB = 30m, OC = 6m, and BC = 50m.

The equation for this parabola will be,

x² = 4ay, which passes through A (50, 30 -6) = (50, 24)

50² = 4a(24)

4a = $\frac{50*50}{24}$

4a = $\frac{625}{6}$

Equation of the parabola becomes,

x² = $\frac{625}{6}$ y

6x² = 652 y

So, at D where x = 18m

6(18)² = 652 y

y = $\frac{6*18*18}{625}$

y = 3.11 m

DE = 3.11 m

DF = DE + EF = 3.11 +6 = 9.11

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

问题4.拱形为半椭圆形。中心宽8 m，高2 m。在离一端1.5 m的点处找到拱门的高度。

解决方案：

According to the given conditions, we can diagrammatically represent

a = 4m

b = 2m 

The equation of the semi – ellipse will be of the from 

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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

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

OA = (4 – 1.5)m = 2.5m

The x – coordinate of point C is 2.5

On substituting x = 2.5, we get,

$\frac{2.5^2}{16} + \frac{y^2}{4}$ = 1

$\frac{6.25}{16} + \frac{y^2}{4}$ = 1

y² = 4 (1 – $\frac{6.25}{16}$)

= 4 ($\frac{9.75}{16}$)

= 2.4375

y = 1.56 (approx.)

So, AC = 1.56 m

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

问题5.一根12厘米长的杆在移动时其末端始终与坐标轴接触。确定杆上P点的轨迹方程，该点距与x轴接触的一端3 cm。

解决方案：

According to the given conditions, we can diagrammatically represent,

Where, let AB be the rod making an angle θ with OX and P(x,y) be the point on it such that

AP = 3cm and AB = 12cm

Then, PB = AB – AP = (12 – 3) = 9cm

From P, draw PQ ⊥ OY and PR ⊥ OX.

In ΔPBQ, 

cos θ = $\frac{PQ}{PB} = \frac{x}{9}$

Sin θ = $\frac{PR}{PA} = \frac{y}{3}$

we know that, sin²θ +cos²θ = 1,

So,

$(\frac{y}{3})^2 + (\frac{x}{9})^2 = 1$

$\frac{x^2}{81} + \frac{y^2}{9} = 1$

Hence, the equation of the locus of point P on the rod is $\mathbf{\frac{x^2}{81} + \frac{y^2}{9} = 1}$

问题6。找到由抛物线的顶点x ² = 12y与其直肠末端相连的线所形成的三角形区域。

解决方案：

According to the given conditions, we can diagrammatically represent,

The equation for this parabola will be,

x² = 4ay = 12y

4a = 12

a = 3 cm

Focus is (0,a) = (0,3)

Now let AB be the latus rectum of the given parabola.

At y = 3, 

x² = 12(3)

x² = 36

x = ±6

So, the coordinates of A are (-6, 3), while the coordinates of B are (6, 3)

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

By using the formula,

Area of ΔOAB = ½ [0(3-3) + (-6)(3-0) + 6(0-3)]

= ½ [(-6) (3) + 6 (-3)]

= ½ [-18-18]

= ½ [-36]

= 18 sq. unit

Hence, area of ΔOAB is 18 sq. unit

问题7：一个跑马场的人注意到，与他两个旗杆的距离之和总为10 m，而旗杆之间的距离为8 m。找到该男子追踪的职位方程式。

解决方案：

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.

So, PA + PB = 10.

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

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

2a = 10

Now let us take the origin of the coordinate plane as the centre of the ellipse, and taking the major axis along the x- axis,

According to the given conditions, we can diagrammatically represent,

The equation for this ellipse will be,

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where ‘a’ is the semi-major axis.

So, 2a = 10

a = 5

Distance between the foci, 

2c = 8

c = 4

By using the relation, we get 

c = √(a² – b²) 

4 = √(25 – b²)

16 = 25 – b²

b² = 25 -16 = 9

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

问题8.在抛物线y ² = 4轴上刻有等边三角形，其中一个顶点在抛物线的顶点。找出三角形边的长度。

解决方案：

According to the given conditions, we can diagrammatically represent,

Let AB intersect the x – axis at point C.

Now let OC = k

The equation for this parabola for x = k,

y² = 4ak

y = ±2√ak

The coordinates of points A and B are (k, 2√ak), and (k, -2√ak)

AB = CA + CB

= 2√ak + 2√ak

= 4√ak

So, OA² = OC² + AC² (Pythagoras theorem)

OA² = k² + (2√ak)²

Since, OAB is an equilateral triangle, OA² = AB².

k² + (2√ak)² = (4√ak)²

k² + 4ak = 16ak

k² = 12ak

k = 12a

So, AB = 4√ak = 4√(a×12a)

= 4√12a²

= 4√(4a²×3)

= 8a√3

Hence, the side of the equilateral triangle inscribed in parabola y² = 4ax is 8a√3 units.