Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10b Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Conic Sections Solutions Exercise 10b
I. In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
Question 1.
y2 = 12z
Solution:
The given equation is y2 = 12x
Here, the coefficient of x is positive, hence the parabola opens towards the right.
On comparing this equation with y2 = 4ax,
we have 4a = 12 ⇒ a = 3
Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y2, the axis of the parabola is the X-axis.
Equation of directrix, x = – a, i.e., x = -3 ⇒ x + 3 = 0
Length of the latus rectum = 4a = 4 × 3 = 12.
Question 2.
x2 = 6y
Solution:
The given equation is x2 = 6y
Here, the coefficient ofy is positive. Hence, the parabola opens upwards on comparing the equation with x2 = 4ay
4a = 6 ⇒ a = \(\frac{3}{2}\)
⇒ Coordinates of the focus = (0, a) = (0, \(\frac{3}{2}\))
Since the given equation involves x2, the axis of the parabola is the Y-axis (x = 0).
Equation of directrix y = \(\frac{-3}{2}\). Length of the latus rectum = 4a = 6
Question 3.
y2 = -8x
Solution:
The given equation is y2 = -8x
Here, the coefficient of “x’ is negative. Hence, the parabola opens towards left. On comparing this equation with y2 = – 4ax
– 4a = -8
⇒ a = 2
Coordinates of the focus = (-a, 0) = (- 2, 0).
Since the given equation involves y2, the axis of the parabola is the X-axis (y = 0). Equation of the directrix, x = a i.e., x = 2, Length of the latus rectum = 4a = 8.
Question 4.
x2 = -16y
Solution:
The given equation is x2 = -16y
Here, the coefficient of y is negative, hence the parabola opens downwards. On comparing this equation with x2 = – 4ay, we have -4a = -16
⇒ a = 4
∴ The coordinates of the focus = (0, -a) = (0, -4)
Since the equation involves x2, the axis of the parabola is Y-axis (x = 0).
Equation of directrix, y = a i.e., y = 4
Length of the latus rectum = 4a = 16
Question 5.
y2 = 10x
Solution:
The given equation is y2 = 10x
Here, the coefficient of x is positive, hence the parabola opens towards the right.
On comparing this equation with y2 = 4ax,
4a = 10 ⇒ a = \(\frac{5}{2}\),
∴ Coordinates of the focus = (a, 0) = (\(\frac{5}{2}\), 0)
Since the given equation involves y2, the axis of the parabola is the X-axis (y = 0).
Equation of the directrix is x = -a i.e., x = \(\frac{-5}{2}\). Length of the latus rectum = 4a = 10
Question 6.
x2 = – 9y
Solution:
The given equation is x2 = – 9y
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x2 = – 4ay,
-4a = – 9
a = \(\frac{9}{4}\)
∴ Coordinates of the focus = (0, -a) = (0, \(\frac{-9}{4}\))
Since, the equation involves x2, the axis of the parabola is the Y-axis (x = 0).
Equation of the directrix y = a i.e., y = \(\frac{9}{4}\). Length of the latus rectum = 4a = 9.
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions.
Question 7.
Focus (6, 0); directrix x = – 6
Solution:
Focus (6, 0); directrix, x = – 6
Since the focus lies on the X-axis, the X-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y2 = 4ax or y2 = -4ax
It is also seen that the directrix, x = – 6 is to the left of the Y-axis, while the focus (6, 0) is right of the Y-axis.
Hence, the parabola is in the form of y2 = 4ax with a = 6
Thus, the equation of the parabola is y2 = 24x.
Question 8.
Focus (0, – 3); directrix y = 3
Solution:
Focus = (0, – 3); directrix y = 3
Since the focus lies on Y-axis, the Y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x2 = 4ay or x2 = -4ay.
It is also seen that the directrix, y = 3 is above the X-axis, while the focus (0, -3) is below the X-axis.
Hence, the parabola is of the form x2 = -4ay with a = 3
Thus, the equation of the parabola is x2 = -12y.
Question 9.
Vertex (0, 0); focus (3, 0)
Solution:
Vertex (0, 0); focus (3, 0)
Since the vertex of the parabola is (0, 0), and the focus lies on the positive X-axis.
Here, X-axis is the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax. Since the focus is (3, 0), a = 3
Thus, the equation of the parabola is y2 = 4 × 3 × x
⇒ y2 = 12x.
Question 10.
Vertex (0, 0); focus (-2, 0)
Solution:
Vertex (0, 0); focus (-2, 0)
Since the vertex of the parabola is (0, 0), and the focus lies on the negative X-axis.
Here, X-axis is the axis of the parabola, while the equation of the parabola is of the
form y2 = 4ax. Since the focus is (-2, 0), a = 2
Thus, the equation of the parabola is y2 = – 4 × —2 × x
⇒ y2 = 8x.
Question 11.
Vertex (0. 0) passing through (2, 3) and axis is along X-axls.
Solution:
Since the vertex is (0, 0) and the axis of the parabola is the X-axis, the equation of the parabola is either of the form y2 = 4ax or y2 = – 4ax.
The parabola passes through the point (2, 3), which lies in the first quadrant.
So, the equation of the parabola is of the form y2 = 4ax, while (2, 3) must satisfy the equation y2 = 4ax,
∴ 32 = 4a(2)
a = \(\frac{9}{8}\)
Thus, the equation of the parabola is y2 = 4\(\left(\frac{9}{8}\right)\)x
⇒ 2y2 = 9x
Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect to Y-axis.
Solution:
Since the vertex is (0, 0) and the parabola is symmetric about the Y-axis, the equation of the parabola is either of the form x2 = 4ay or x2 = – 4ay.
The parabola passes through the point (5, 2), which lies in the first quadrant.
So, the equation of the parabola is of the form x2 = 4ay, while (5, 2) must satisfy the equation x2 = 4ay
∴ (5)2 = 4 × a × 2
⇒ 25 = 8a
⇒ a = \(\frac{25}{8}\)
II.
Question 1.
Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).
Solution:
Given S = (3, 5), A =(1, 3).
Equation of the axis is y – 3 = \(\frac{3-5}{1-3}\)(x – 1)
⇒ y – 3 = x – 1
⇒ x – y + 2 = 0
The directrix is perpendicular to axis. Equation of the directrix isx + y + k = 0
Let the co-ordinates of Z be (x, y). A is the midpoint of SZ. Then,
\(\frac{3+x}{2}\) = 1 ⇒ 3 + x = 2
⇒ x = -1; \(\frac{5+y}{2}\) = 3
⇒ 5 + y = 6
⇒ y = 1.
Co-ordinates of Z are (-1, 1).
The directrix passes through Z (-1, 1) ⇒ -1 + 1 + k = 0 ⇒ k = 0
Equation of the directrix is x + y = 0
Equation of the parabola is (x – α)2 + (y + β)2 = \(\frac{(l x+m y+n)^2}{l^2+m^2}\)
(x – 3)2 + (y – 5)2 = \(\frac{(x+y)^2}{1+1}\)
⇒ 2(x2 – 6x + 9 + y2 – 10y + 25) = (x + y)2
⇒ 2x2 + 2y2 – 12x – 20y + 68 = x2 + 2xy + y2
⇒ x2 – 2xy + y2 – 12x – 20y + 68 = 0
Question 2.
Find the equation of the parabola whose focus is (-2, 3) and directrix is the line 2x + 3y – 4 = 0. Also find the length of the latus rectum and the equation of the axis of the parabola.
Solution:
Assume P (x1, y1) is any point on the parabola, S(-2, 3) is the focus.
⇒ SP2 = (x1 + 2)2 + (y1 – 3)2.
Equation of the directrix is 2x + 3y – 4 = 0.
PM is the perpendicular from P on the directrix PM = \(\frac{\left|2 x_1+3 y_1-4\right|}{\sqrt{4+9}}\)
SP = PM
⇒ SP2 = PM2
⇒ (x1 + 2)2 + (y1 – 3)2 = \(\frac{1}{2}\)
⇒ 13(x12 + 4x1 + 4 + y12 – 6y1 + 9) = (2x1 + 3y1 – 4)2
⇒ 13x12 + 13y12 + 52x1 – 78y1 + 169) = 4x12 + 9y12 + 16 + 12x1y1 – 16x1 – 24y1
⇒ 9x12 – 12x1y1 + 4y12 + 68x1 – 54y1 + 153 = 0
Locus of P(x1, y1) is 9x2 – 12xy + 4y2 + 68x – 54y + 153 = 0
Length of the latus rectum = 4a
Perpendicular distance from S on directrix = \(\frac{|2(-2)+3(3)-4|}{\sqrt{4+9}}=\frac{1}{\sqrt{13}}\)
2a = \(\frac{1}{\sqrt{13}}\) ⇒ 4a = \(\frac{2}{\sqrt{13}}\). The axis is perpendicular to the directrix.
Equation of the directrix can be taken as 3x – 2y + k = 0.
The line passes through S (-2, 3) ⇒ -6 – 6 + k = 0
⇒ k = 12
Equation of the axis is 3x – 2y + 12 = 0.
Question 3.
Find the equation of the parabola whose axis is parallel to X-axis and which passes through the points (-2, 1), (1, 2) and (-1, 3).
Solution:
Axis is parallel to X – axis, x = ay2 + by + c.
Passes through (-2, 1), (1, 2), (-1, 3)
-2 = a + b + c ………… (1)
1 = 4a + 2b + c ……….. (2)
– 1 = 9a + 3b + c ……… (3)
(2) – (3), we get 2 = – 5a – b …….. (4)
(2) – (1), we get 3 = 3a + b …… (5)
(4) + (5), we get 5 = – 2a ⇒ a = \(\frac{-5}{2}\) ⇒ b = \(\frac{21}{2}\),c = -10
x = \(\frac{-5}{2}\)y2 + \(\frac{21}{2}\)y – 10
⇒ 5y2 + 2x – 21y + 20 = 0.
Question 4.
Find the equation of the parabola whose axis is parallel to Y-axis and which passes through the points (4, 5), (-2, 11) and (-4, 21).
Solution:
Axis is parallel to Y-axis, y = ax2 + bx + c.
Passes through (4, 5), (-2, 11), (-4, 21)
5 = 16a + 4b + c ……….. (1)
11 = 4a – 2b + c ……….. (2)
21 = 16a – 4b + c ……..(3)
(2) – (1), we get 6 = -12a – 6b ….. (4)
(2) – (3), we get 10 = 12a – 2b ……. (5)
Solve (4) and (5), we get a = \(\frac{1}{2}\)
⇒ b = -2, c = 5
y = \(\frac{1}{2}\)x2 – 2x + 5
⇒ x2 – 2y – 4x + 10 = 0.