## NCERT Solutions Class 11th Maths Chapter 11: Conic Sections## Exercise 11.1
Centre of the circle = (0, 2) Radius of the circle = 2 units We know that the equation of a circle with centre (h, k) and radius r is given by (x - h) Therefore, (x - 0) x x Hence, the equation of the circle is x
Centre of the circle = (-2, 3) Radius of the circle = 4 units We know that the equation of a circle with centre (h, k) and radius r is given by (x - h) Therefore, (x + 2) x x x Hence, the equation of the circle is x
Centre of the circle = (1/2, 1/4) Radius of the circle = 1/12 units We know that the equation of a circle with centre (h, k) and radius r is given by (x - h) Therefore, (x - 1/2) x 144(x 144x 144x 4(36x 36x Hence, the equation of the circle is 36x
Centre of the circle = (1, 1) Radius of the circle = √2 units We know that the equation of a circle with centre (h, k) and radius r is given by (x - h) Therefore, (x - 1) x x Hence, the equation of the circle is x
Centre of the circle = (-a, -b) Radius of the circle = √(a We know that the equation of a circle with centre (h, k) and radius r is given by (x - h) Therefore, (x + a) x x Hence, the equation of the circle is x
6. (x + 5)
Equation of the circle is (x + 5) Compare the given equation to (x - h) Therefore, h = -5 k = 3 r r = 6 Hence, the centre of given circle is (-5, 3) and its radius is 6 units.
Equation of the circle is x Add 4 and 16 to both sides x x (x - 2) Compare the given equation to (x - h) Therefore, h = 2 k = 4 r r = √65 Hence, the centre of given circle is (2, 4) and its radius is √65 units.
Equation of the circle is x Add 16 and 25 to both sides x x (x - 4) Compare the given equation to (x - h) Therefore, h = 4 k = -5 r r = √53 Hence, the centre of given circle is (4, -5) and its radius is √53 units.
Equation of the circle is 2x 2(x x Add 1/16 to both sides x x (x - 1/4)
Therefore, h = 1/4 k = 0 r r = 1/4 Hence, the centre of given circle is (1/4, 0) and its radius is 1/4 units.
Let the equation of required circle be (x - h) It is given that (4, 1) and (6, 5) lie on the circle. Therefore, (4 - h) AND (6 - h) So, (4 - h) 16 + h 4h + 8k = 44 4(h + 2k) = 44
h = 11 - 2k It is also given that the centre of the circle lies on the line 4x + y = 16. Therefore, 4h + k = 16 Substitute h = 11 - 2k 4(11 - 2k) + k = 16 44 - 8k + k = 16 28 = 7k
So, h = 11 - 8
Thus, the centre of circle is (3, 4). (4 - h) (4 - 3) 1 + 9 = r r
Thus, radius of the circle is √10. Hence, the equation of the required circle is (x - 3)
Let the equation of required circle be (x - h) It is given that (2, 3) and (-1, 1) lie on the circle. Therefore, (2 - h) AND (-1 - h) So, (2 - h) 4 + h
6h = 11 - 4k h = (11 - 4K)/6 It is also given that the centre of the circle lies on the line x - 3y - 11 = 0. Therefore, h - 3k = 11 Substitute h = (11 - 4k)/6 (11 - 4k)/6 = 11 + 3k 11 - 4k = 66 + 18k 22k = -55
So, h = (11 + 10)/6
Thus, the centre of circle is (7/2, -5/2). (2 - h) (2 - 7/2) (4 - 7) 9/4 + 121/4 = r r r
Thus, radius of the circle is √(65/2). Hence, the equation of the required circle is (x - 7/2)
Let the equation of required circle be (x - h) The centre lies on the x-axis, so k = 0. Radius of the circle, r = 5 units Equation of the circle will be (x - h) It is given that the point (2, 3) lies on the circle. Therefore, (2 - h) (2 - h) (2 - h) (2 - h) Square root both sides (2 - h) = ± 4 If (2 - h) = 4 ⇒ If (2 - h) = -4 ⇒ CASE I: h = -2 Equation of the circle will be (x + 2) x
CASE II: h = 6 Equation of the circle will be (x - 6) x
Hence, the equation of the required circle is x
It is given that the point (0, 0) lies on the circle. Therefore, (0 - h) h So, (x - h) It is also given that the circle intercepts a and b on the axes. Therefore, it passes through (a, 0) and (0, b). (a - h) AND (0 - h) Taking (a - h) a a a(a - 2h) = 0 Since, a ≠ 0. a - 2h = 0
Taking (0 - h) h b b(b - 2k) = 0 Since, b ≠ 0 b - 2k = 0
Thus, equation of the circle is (x - a/2) (2x - a) 1/4 × [4x 4x 4x 4(x
Hence, the equation of the required circle is x
Centre of the circle is (h, k) = (2, 2) Radius of the circle, r = Distance between the centre and one point on the circle r = √[(2 - 4) = √(4 + 9) = √13 Thus, equation of the circle will be (x - h) (x - h) (x - 2) x
Hence, the equation of the required circle is x
Equation of the given circle is x (x - 0)
Therefore, h = 0 k = 0 r r = √25 = 5 Distance of the given point (-2.5, 3.5) from the centre of circle (0, 0) is d = √[(-2.5 + 0) = √(6.25 + 12.25) = √18.5 d < r Hence, the given point lies inside the circle x ## Exercise 11.2
y Coefficient of x = 12 We know that the parabola opens towards the right when coefficient of x is positive. Compare the given equation to y 4a = 12
Coordinates of the focus of parabola = (a, 0) = (3, 0) The given equation involves y Equation of the directrix is x = -a x + a = 0 x + 3 = 0 Length of the latus rectum = 4a = 4(3) = 12
x Coefficient of y = 6 We know that the parabola opens upwards when coefficient of y is positive. Compare the given equation to x 4a = 6
Coordinates of the focus of parabola = (0, a) = (0, 3/2) The given equation involves x Equation of the directrix is y = -a y + a = 0 y + 3/2 = 0 Length of the latus rectum = 4a = 4(3/2) = 6
y Coefficient of x = -8 We know that the parabola opens towards the left when coefficient of x is negative. Compare the given equation to y 4a = 8
Coordinates of the focus of parabola = (-a, 0) = (-2, 0) The given equation involves y Equation of the directrix is x = a x - a = 0 x - 2 = 0 Length of the latus rectum = 4a = 4(2) = 8
x Coefficient of y = -16 We know that the parabola opens downwards when coefficient of y is negative. Compare the given equation to x 4a = 16
Coordinates of the focus of parabola = (0, -a) = (0, -4) The given equation involves x Equation of the directrix is y = a y - a = 0 y - 4 = 0 Length of the latus rectum = 4a = 4(4) = 16
y Coefficient of x = 0 We know that the parabola opens towards the right when coefficient of x is positive. Compare the given equation to y 4a = 10
Coordinates of the focus of parabola = (a, 0) = (5/2, 0) The given equation involves y Equation of the directrix is x = -a x + a = 0 x + 5/2 = 0 Length of the latus rectum = 4a = 4(5/2) = 10
x Coefficient of y = -9 We know that the parabola opens downwards when coefficient of y is negative. Compare the given equation to x 4a = 9
Coordinates of the focus of parabola = (0, -a) = (0, -9/4) The given equation involves x Equation of the directrix is y = a y - a = 0 y - 9/4 = 0 Length of the latus rectum = 4a = 4(9/4) = 9
The focus is at (6, 0) which lies on the x-axis. So, the axis of the parabola is x-axis. Therefore, the equation of required parabola is either in the form of y Directrix of the parabola is -6, which is to the left of y-axis and the focus is to the right of y-axis. Thus, the required parabola is of the form Focus is given by (a, 0). (6, 0) = (a, 0) which implies that Hence, the equation of the parabola is y
The focus is at (0, -3) which lies on the y-axis. So, the axis of the parabola is y-axis. Therefore, the equation of required parabola is either in the form of x Directrix of the parabola is 3, which is above the x-axis and the focus is below the x-axis. Thus, the required parabola is of the form Focus is given by (0, a). (0, -3) = (0, a) which implies that Hence, the equation of the parabola is x
The focus is at (3, 0) which lies on the positive x-axis. So, the axis of the parabola is x-axis. The vertex of the parabola is at (0, 0). Thus, the required parabola is of the form y Focus is given by (a, 0). (3, 0) = (a, 0) which implies that Hence, the equation of the parabola is y
The focus is at (-2, 0) which lies on the negative x-axis. So, the axis of the parabola is x-axis. The vertex of the parabola is at (0, 0). Thus, the required parabola is of the form y Focus is given by (-a, 0). (-2, 0) = (-a, 0) which implies that Hence, the equation of the parabola is y
It is given that the vertex of the parabola is (0, 0) and its axis is along the x-axis. So, the equation of the parabola is of the form y It is also given that the parabola passes through the point (2, 3) which is a point in the first quadrant. Therefore, The equation of parabola is of the form y 3 9 = 8a
Equation of parabola will be y y Hence, the equation of the parabola is 2y
It is given that the vertex of the parabola is (0, 0) and it is symmetric with respect to y-axis. So, the equation of the parabola is of the form x It is also given that the parabola passes through the point (5, 2) which is a point in the first quadrant. Therefore, The equation of parabola is of the form x 5 25 = 8a
Equation of parabola will be x x Hence, the equation of the parabola is 2x ## Exercise 11.3
In x On comparing the given equation to x a AND b c = √(a = √(36 - 16) = √20 = 2√5 Now, Coordinates of the foci = (2√5, 0) and (-2√5, 0). Coordinates of the vertices = (6, 0) and (-6, 0) Length of the major axis = 2(6) = 12 units Length of the minor axis = 2(4) = 8 units Eccentricity = e = 2√5/6 = √5/3 Length of the latus rectum = 2b
/25 = 1
In x On comparing the given equation to x a AND b c = √(a = √(25 - 4) = √21 Now, Coordinates of the foci = (0, √21) and (0, -√21). Coordinates of the vertices = (0, 5) and (0, -5) Length of the major axis = 2(5) = 10 units Length of the minor axis = 2(2) = 4 units Eccentricity = e = √21/5 Length of the latus rectum = 2b
In x On comparing the given equation to x a AND b c = √(a = √(16 - 9) = √7 Now, Coordinates of the foci = (√7, 0) and (-√7, 0). Coordinates of the vertices = (4, 0) and (-4, 0) Length of the major axis = 2(4) = 8 units Length of the minor axis = 2(3) = 6 units Eccentricity = e = √7/4 Length of the latus rectum = 2b
In x On comparing the given equation to x a AND b c = √(a = √(100 - 25) = √75 = 5√3 Now, Coordinates of the foci = (0, 5√3) and (0, -5√3). Coordinates of the vertices = (0, 10) and (0, -10) Length of the major axis = 2(10) = 20 units Length of the minor axis = 2(5) = 10 units Eccentricity = e = 5√3/10 = √3/2 Length of the latus rectum = 2b
In x On comparing the given equation to x a AND b c = √(a = √(49 - 36) = √13 Now, Coordinates of the foci = (√13, 0) and (-√13, 0). Coordinates of the vertices = (7, 0) and (-7, 0) Length of the major axis = 2(7) = 14 units Length of the minor axis = 2(6) = 12 units Eccentricity = e = √13/7 Length of the latus rectum = 2b 6. x
In x On comparing the given equation to x a AND b c = √(a = √(400 - 100) = √300 = 10√3 Now, Coordinates of the foci = (0, 10√3) and (0, -10√3). Coordinates of the vertices = (0, 20) and (0, -20) Length of the major axis = 2(20) = 40 units Length of the minor axis = 2(10) = 20 units Eccentricity = e = 10√3/20 = √3/2 Length of the latus rectum = 2b
36x Divide both sides by 144 36x x In x On comparing the given equation to x a AND b c = √(a = √(36 - 4) = √32 = 4√2 Now, Coordinates of the foci = (0, 4√2) and (0, -4√2). Coordinates of the vertices = (0, 6) and (0, -6) Length of the major axis = 2(6) = 12 units Length of the minor axis = 2(2) = 4 units Eccentricity = e = 4√2/6 = 2√2/3 Length of the latus rectum = 2b
16x Divide both sides by 16 16x x In x On comparing the given equation to x a AND b c = √(a = √(16 - 1) = √15 Now, Coordinates of the foci = (0, √15) and (0, -√15). Coordinates of the vertices = (0, 4) and (0, -4) Length of the major axis = 2(4) = 8 units Length of the minor axis = 2(1) = 2 units Eccentricity = e = √15/4 Length of the latus rectum = 2b
4x Divide both sides by 36 4x x In x On comparing the given equation to x a AND b c = √(a = √(9 - 4) = √5 Now, Coordinates of the foci = (√5, 0) and (-√5, 0). Coordinates of the vertices = (3, 0) and (-3, 0) Length of the major axis = 2(3) = 6 units Length of the minor axis = 2(2) = 4 units Eccentricity = e = √5/3 Length of the latus rectum = 2b
It is given that the vertices are (± 5, 0) which lie on the x-axis. Therefore, the equation of the ellipse will be of the form x a = 5 and c = 4 We know that a So, 5 b b b = 3 Hence, the equation of the ellipse will be x
It is given that the vertices are (0, ± 13) which lie on the y-axis. Therefore, the equation of the ellipse will be of the form x a = 13 and c = 5 We know that a So, 13 b b b = 12 Hence, the equation of the ellipse will be x
It is given that the vertices are (± 6, 0) which lie on the x-axis. Therefore, the equation of the ellipse will be of the form x a = 6 and c = 4 We know that a So, 6 b b b = 2√5 Hence, the equation of the ellipse will be x
Ends of major axis are (± 3, 0) which is along the x-axis. So, the equation of the ellipse will be of the form x a = 3 and b = 2 a Hence, the equation of the ellipse will be x
Ends of major axis are (0, ± √5) which is along the y-axis. So, the equation of the ellipse will be of the form x a = √5 and b = 1 a Hence, the equation of the ellipse will be x
The given foci is (± 5, 0) which lies on the x-axis. Therefore, x-axis is the major axis. So, the equation of the ellipse will be of the form x Length of major axis = 2a 2a = 26 a = 13 and c = 5 We know that a So, 13 b b b = 12 Hence, the equation of the ellipse will be x
The given foci is (0, ± 6) which lies on the y-axis. Therefore, y-axis is the major axis. So, the equation of the ellipse will be of the form x Length of minor axis = 2b 2b = 16 b = 8 and c = 6 We know that a So, a a a a = 10 Hence, the equation of the ellipse will be x
The given foci is (± 3, 0) which lies on the x-axis. Therefore, x-axis is the major axis. So, the equation of the ellipse will be of the form x a = 4 and c = 3 We know that a So, 4 b b b = √7 Hence, the equation of the ellipse will be x
It is given that the foci are on the x-axis. Therefore, x-axis is the major axis. So, the equation of the ellipse will be of the form x b = 3 and c = 4 We know that a So, a a a a = 5 Hence, the equation of the ellipse will be x
It is given that the centre is at (0, 0) and the major axis is on the y- axis. Therefore, the equation of ellipse will be of the form x It is also given that the ellipse passes through the point (3, 2) and (1, 6). Therefore, 3 9/b AND 1 1/b 1/b 1/b Substitute the obtained value of 1/b 9 × (a (9a 9a 8a
So, 1/b 1/b ⇒ Hence, the equation of the ellipse will be x
It is given that the major axis is on the x-axis. Therefore, the equation of ellipse will be of the form x It is also given that the ellipse passes through the point (4, 3) and (6, 2). Therefore, 4 16/a AND 6 36/a 4/b 1/b Substitute the obtained value of 1/b 16/a (64 + 9a 9a 5a
So, 1/b 1/b 1/b ⇒ ## Exercise 11.4
x x On comparing the given equation x a = 4 b = 3 We know that a 4 16 + 9 = c 25 = c c = 5 Therefore, Coordinates of the foci are (5, 0) and (-5, 0) Coordinates of the vertices are (4, 0) and (-4, 0) Eccentricity = e = c/a = 5/4 Length of the latus rectum = 2b
y y On comparing the given equation y a = 3 b = √27 We know that a 3 9 + 27 = c 36 = c c = 6 Therefore, Coordinates of the foci are (0, 6) and (0, -6) Coordinates of the vertices are (0, 3) and (0, -3) Eccentricity = e = c/a = 6/3 = 2 Length of the latus rectum = 2b
9y Divide both sides by 36 9y y y On comparing the given equation y a = 2 b = 3 We know that a 2 4 + 9 = c 13 = c c = √13 Therefore, Coordinates of the foci are (0, √13) and (0, -√13) Coordinates of the vertices are (0, 2) and (0, -2) Eccentricity = e = c/a = √13/2 Length of the latus rectum = 2b
16x Divide both sides by 576 16x x x On comparing the given equation x a = 6 b = 8 We know that a 6 36 + 64 = c 100 = c c = 10 Therefore, Coordinates of the foci are (10, 0) and (-10, 0) Coordinates of the vertices are (6, 0) and (-6, 0) Eccentricity = e = c/a = 10/6 = 5/3 Length of the latus rectum = 2b
5y Divide both sides by 36 5y y y On comparing the given equation y a = 6/√5 b = 2 We know that a (6/√5) 36/5 + 4 = c (36 + 20)/5 = c 56/5 = c c = 2√14/√5 Therefore, Coordinates of the foci are (0, 2√14/√5) and (0, -2√14/√5) Coordinates of the vertices are (0, 6/√5) and (0, -6/√5) Eccentricity = e = c/a = (2√14/√5)/(6/√5) = √14/3 Length of the latus rectum = 2b
49y Divide both sides by 784 49y y y On comparing the given equation y a = 4 b = 7 We know that a (4) 16 + 49 = c 65 = c c = √65 Therefore, Coordinates of the foci are (0, √65) and (0, -√65) Coordinates of the vertices are (0, 4) and (0, -4) Eccentricity = e = c/a = √65/4 Length of the latus rectum = 2b
It is given that the vertices are (± 2, 0) which lie on the x-axis. Therefore, the equation of the hyperbola will be of the form x a = 2 and c = 3 We know that a So, 2 b b b = √5 Hence, the equation of the hyperbola is x
It is given that the vertices are (0, ± 5) which lie on the y-axis. Therefore, the equation of the hyperbola will be of the form y a = 5 and c = 8 We know that a So, 5 b b b = √39 Hence, the equation of the hyperbola is y
It is given that the vertices are (0, ± 3) which lie on the y-axis. Therefore, the equation of the hyperbola will be of the form y a = 3 and c = 5 We know that a So, 3 b b b = 4 Hence, the equation of the hyperbola is y
The given foci is (± 5, 0) which lies on the x-axis. Therefore, the equation of the hyperbola will be of the form x c = 5 Length of the transverse axis = 8 2a = 8 a = 4 We know that a 4 b b b = 3 Hence, the equation of the hyperbola is x
The given foci is (0, ± 13) which lies on the y-axis. Therefore, the equation of the hyperbola will be of the form y c = 13 Length of the conjugate axis = 24 2b = 24 b = 12 We know that a a a a a = 5 Hence, the equation of the hyperbola is y
The given foci is (± 3√5, 0) which lies on the x-axis. Therefore, the equation of the hyperbola will be of the form x c = 3√5 Length of the latus rectum = 8 2b 2b b b We know that a a a a a a(a - 5) + 9(a - 5) = 0 (a - 5)(a + 9) = 0 (a - 5) = 0 ⇒ OR (a + 9) = 0 ⇒ Since, a is not negative, a = -9 is rejected. Therefore, a = 5. So, b b b = 2√5 Hence, the equation of the hyperbola is x
The given foci is (± 4, 0) which lies on the x-axis. Therefore, the equation of the hyperbola will be of the form x c = 4 Length of the latus rectum = 12 2b 2b b b We know that a a a a a a(a - 2) + 8(a - 2) = 0 (a - 2)(a + 8) = 0 (a - 2) = 0 ⇒ OR (a + 8) = 0 ⇒ Since, a is not negative, a = -8 is rejected. Therefore, a = 2. So, b b b = 2√3 Hence, the equation of the hyperbola is x
It is given that the vertices are (± 7, 0) which lie on the x-axis. Therefore, the equation of the hyperbola will be of the form x a = 7 Eccentricity, e = 4/3 c/a = 4/3 c/7 = 4/3 c = 28/3 We know that a So, 7 b b b b = 7√7/3 Hence, the equation of the hyperbola is x
The given foci is (0, ± √10) which lies on the y-axis. Therefore, the equation of the hyperbola will be of the form y c = √10 We know that a So, b
It is given that the hyperbola passes through the point (2, 3). Therefore, (2, 3) will satisfy the equation of the hyperbola 3 9/a Substitute the obtained value of b 9/a [9(10 - a 90 - 9a 90 - 13a a a a (a a OR a We know that in a hyperbola, c > a. So, c Therefore, a Then, b Hence, the equation of the hyperbola is y ## Miscellaneous Exercise
The origin of the coordinate plane is taken as the vertex of the parabolic reflector, where the axis of the reflector is along the positive x-axis. Since, the parabola is opening to the right. Its equation will be of the form y Radius of the parabola = 20/2 = 10 cm Depth of the parabola = 5 cm So, the parabola passes through the point (5, 10). Therefore, 10 100 = 20a a = 5 Focus of the parabola = (a, 0) = (5, 0) (5, 0) is at the mid-point of the diameter. Hence, the focus of the reflector is at the mid-point of the diameter.
The origin of the coordinate plane is taken as the vertex of the parabolic arc, where the axis of the arc is along the positive y-axis. Since, the parabola is opening upwards. Its equation will be of the form x Radius of the parabola = 5/2 m Height of the arch = 10 m So, the parabola passes through the point (5/2, 10). Therefore, (5/2) 25/4 = 40a a = 5/32 The equation of the parabola will be x When y = 2, we get x x x = √(5/4) x = √5/2 Width of the parabola when it is 2 m from the vertex = 2 × √5/2 = √5 m ≈ 2.23 m. Hence, it is 2.23 m wide when its 2 m from the vertex of the parabola.
The origin of the coordinate plane is taken as the vertex of the parabola, where the vertical axis is along the positive y-axis. Longest wire attached to the cable = AB = 30 m Shortest wire attached to the cable = OC = 6 m DF is the supporting wire attached to the roadways at a distance of 18 m from the middle. BC = 50 m The parabola is opening upwards. So, its equation will be of the form x Distance of point A from x-axis = 30 - 6 = 24 m Distance of point A from y-axis = 100/2 = 50 m The point A (50, 24) lies on the parabola. Therefore, it will satisfy the equation 50 2500 = 96a a = 625/24 The equation of the parabola will be x When x = 18, we get 18 y = 1944/625 ≈ 3.11 Therefore, length of DE = 3.11 m Length of DF = DE + EF = 3.11 m + 6 m = 9.11 m Hence, the length of a supporting wire attached to the roadway 18 m from the middle is 9.11 m.
Length of the major axis = Width of semi-ellipse = 8 m = 2a Length of the minor axis = Height of semi-ellipse = 2 m = b The origin is taken as the centre of the semi-ellipse and x-axis is the major axis. So, the equation of the semi-ellipse will be of the form x x Let point A be 1.5 m away from the end point B of the semi-ellipse Draw AC ⊥ OB OA = OB - AB = 4 - 1.5 = 2.5 m x-coordinate of the point C is 2.5 and it lies on the ellipse. Therefore, (2.5) 6.25/16 + y (6.25 + 4y 4y 4y y y ≈ 1.56 Therefore, AC = 1.56 m Hence, the height of the arch at a point 1.5 m from one end is 1.56 m.
Let AB be the rod making the angle θ with x-axis with a point P so that AP = 3 cm. Length of PB = AB - AP = 12 - 3 = 9 cm Draw PQ ⊥ OB and PR ⊥ OA In ∆ BPQ, cos θ = PQ/PB = x/9 sin θ = PR/PA = y/3 sin x Hence, the equation of the locus of a point P on the rod is x
x On comparing the equation of given parabola to x 4a = 12 a = 3 Coordinates of foci = (0, 3) When y = 3, x x x = ± 6 So, the vertices of required triangle will be (0, 0), (6, 3) and (-6, 3). Area of the triangle = 1/2 × [0(3 - 3) + 6(3 - 0) - 6(0 - 3)] = 1/2 × [0 + 18 + 18] = 18 square units
Let the positions of flag posts be A and B and position of the man be P (x, y). Given, PA + PB = 10 m The path followed by the man will be an ellipse as the sum of his distance from two fixed point is constant. A and B will be the foci of this ellipse. The origin is taken as the centre of the ellipse and x-axis is the major axis. So, the equation of the ellipse will be of the form x 2a = PA + PB = 10 a = 5 Distance between foci = 2c = 8 c = 4 We know that c = √(a 4 = √(25 - b 16 = 25 - b b b = 3 Hence, the equation of the man's path is x
The origin O is taken as the vertex of the parabola. Let A and B be the other two vertices of the triangle. AB intersects x-axis at the point C. y Let x = k y y = ± 2√ak So, A and B will be (k, 2√ak) and (k, -2√ak) respectively. Length of AB = CA + CB = 2√ak + 2√ak = 4√ak ∆ OAB is equilateral. Therefore, AO = AB AO OC k k k k = 12a Therefore, AB = 4√(a)(12a) = 4√12a = 4a√12 = 8√3a Hence, the side of the equilateral triangle inscribed in parabola y |