第 11 类 RD Sharma 解决方案 – 第 25 章抛物线 – 练习 25.2
问题 1. 求抛物线 y 2 = x 的对称轴。
解决方案:
We are given,
=> y2 = x
We know this parabola lies on the y-axis.
Therefore, its vertex lies on the y-axis.
Now we know, parabola always has one axis which maintains its symmetry.
In this case, the axis is an x-axis, which divides the parabola into two equal parts.
Hence the required axis is x-axis.
问题2.求抛物线的顶点到焦点的距离y 2 + 6y + 2x + 5 = 0。
解决方案:
Given that,
y2 + 6y + 2x + 5 = 0
(y + 3)2 + 2x – 4 = 0
(y + 3)2 = -2 (x – 2)
Let us assume Y = y + 3 and X = x – 2.
From the equation, we get,
Y2 = – 2X
On putting 4a = 2, we get
=> a = 1/2
Focus = (X = -1/2, Y = 0) = (x = 3/2, y = -3)
Vertex = (X = 0, Y = 0) = (x = 2, y = -3)
Thus, we get
Focus = (3/2, -3)
Vertex = (2, -3)
The distance between the vertex and the focus is,
D =
=
= 1/2 units
Therefore, the required distance is 1/2 units.
问题 3. 求抛物线 x 2 − 4x − 8y + 12 = 0 的准线方程。
解决方案:
Given that,
x2 − 4x − 8y + 12 = 0
(x – 2)2 – 4 – 8y + 12 = 0
(x – 2)2 = 8 (y – 1)
Let Y = y − 1 and X = x – 2.
From the above equation, we get,
X2 = 8Y
On comparing with x2 = 4ay, we get
=> a = 2
Directrix = Y = −a
=> y − 1 = −a
=> y = −a + 1
=> y = −2 + 1
=> y = −1
Therefore, the equation of the directrix is y = -1.
问题 4. 写出焦点 (0, 0) 和准线 x + y − 4 = 0 的抛物线方程。
解决方案:
Let us assume P (x, y) be any point on the parabola whose focus is S (0, 0).
And the equation of directrix is x + y= 4.
First we draw PM perpendicular to x + y = 4.
Then, we have,
SP = PM
SP2 = PM2
(x – 0)2 + (y – 0)2 =
x2 + y2 =
2x2 + 2y2 = x2 + y2 + 16 + 2xy – 8y – 8x
=> x2 + y2 – 2xy + 8x + 8y – 16 = 0
Hence, the equation of the parabola is x2 + y2 – 2xy + 8x + 8y – 16 = 0
问题 5. 求抛物线 y 2 = 4ax 的弦长穿过顶点并以 π/4 倾斜于轴。
解决方案:
Let us considered OP be the chord and the coordinates of P be (x1, y1).
We have,
=> OP2 = x12 + y12
And, tan π/4 = y1/x1
=> 1 = y1/x1
=> x1 = y1
and (x1, y1) lies on the parabola.
∴ y12 = 4ax1
So, we get
=> x12 = 4a x1
=> x1 = 4a
Therefore, we have
OP2 = (4a)2 + (4a)2
OP2 = 32a2
OP = 4√2 a
Therefore, the length of the chord is 4√2 a.
问题 6. 如果 b 和 c 是抛物线 y 2 = 4ax 的任意焦弦段的长度,则写出其直角直肠的长度。
解决方案:
Let us assume S (a, 0) be the focus of the given parabola and the end points of the focal chord be P(at2 , 2at) and Q (a/t2}, -2a/t).
Now, SP and SQ are segments of the focal chord with lengths b and c.
∴ SP = b, SQ = c
Also, SP =
= a (1 + t2)
And, SQ =
= a (1 + 1/t2)
Now, we get,
=> a =
Hencec, the length of the latus rectum is 4a =
问题 7. PSQ 是抛物线 y 2 = 8x 的焦弦。如果 SP = 6,则求 SQ。
解决方案:
According to the question, it is given that PSQ is a focal chord of the parabola y2 = 8x and SP = 6
So, the coordinates of the focal chord are P (at2 , 2at) and Q (a/t2, – 2a/t).
On comparing y2 = 8x with y2 = 4ax, we get,
=> 4a = 8
=> a = 2
Therefore, the coordinates of the focus S is (2, 0).
SP = 6
t4 + 2t2 – 8 = 0
=> t2 = 2
Thus, we get,
SQ =
=
=
= √5
Hence, SQ is √5.
问题 8. 写出焦点在 (-2, 1) 的抛物线顶点的坐标,准线是直线 x + y - 3 = 0。
解决方案:
Given that, the focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0 and the slope of the line perpendicular to x + y − 3 = 0 is 1.
Now, the axis of the parabola is perpendicular to the directrix and passes through the focus.
So, the equation of the axis of the parabola becomes,
=> y – 1 = 1 (x + 2)
Now, the intersection point of the directrix and axis is the intersection point of parabola and x + y − 3 = 0.
Let us assume that the intersection point be K.
So, the coordinates of K are (0, 3).
Let us assume that the point (h, k) be the coordinates of the vertex, which is the mid-point of the line segment joining K and the focus.
=> h = (0 – 2)/2, k = (3 + 1)/2
=> h = -1, k = 2
Hence, the coordinates of the vertex are (−1, 2).
问题 9. 如果抛物线的顶点坐标和焦点坐标分别为 (-1, 1) 和 (2, 3),则写出其准线方程。
解决方案:
Given that, the vertex and the focus of a parabola are (−1, 1) and (2, 3).
So, the slope of the axis of the parabola = = 2/3
Slope of the directrix = -3/2
Let us assume the directrix intersect the axis at point K (r, s).
So, (r + 2)/2 = -1, (s + 3)/2 = 1
=> r = -4, s = -1
Now, the required equation of the directrix is,
y + 1 = (-3/2) (x + 4)
=> 3x + 2y + 14 = 0
问题 10. 如果抛物线 y 2 = 4ax 过点 (3, 2),则求其直角的长度。
解决方案:
Given that, y2 = 4ax and the parabola is passing through the point (3, 2)
So, this will satisfy the equation of the parabola.
22 = 4(a)(3)
=> a = 1/3
The length of the latus ractum = 4a
= 4 (1/3)
= 4/3
Hence, the latus ractum is 4/3
问题 11. 写出顶点在 (−3, 0) 且准线为 x + 5 = 0 的抛物线方程。
解决方案:
Given that the vertex(h, k) is (-3, 0) and the directrix is x + 5 = 0
So, the general equation of the parabola = (y − k)2 = 4a (x − h).
Now, the directrix is
x = h − a
As x + 5 = 0, we get
=> −5 = −3 − a
=> a = 2
So, the equation of the parabola is
(y − 0)2 = 4 (2) (x + 3)
=> y2 = 8 (x + 3)