(ii) What is the focal distance of the point (4, 6) lies on the parabola y²=9x ?
Answers
Step-by-step explanation:
Given, y^2 = 9xy
2
=9x
Comparing with standard equation of parabola : y^2 = 4axy
2
=4ax
4a = 9 or a = \frac {9}{4}
4
9
hence, Value for a = \frac {9}{4}a=
4
9
Therefore, Focus = s(\frac {9}{4}
4
9
, 0)
As we know the coordinates of end points of diameter, the equation of circle can be given by:
(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0(x−x
1
)(x−x
2
)+(y−y
1
)(y−y
2
)=0
⇒ (x-4)(x-\frac {9}{4}) + (y-6)(y-0) = 0(x−4)(x−
4
9
)+(y−6)(y−0)=0
⇒ (x-4)(4x-9) + 4y(y-6) = 0(x−4)(4x−9)+4y(y−6)=0
⇒ 4x^2 + 4y^2-25x-24y+36 = 04x
2
+4y
2
−25x−24y+36=0
⇒ x^2 + y^2 - \frac {25}{4}x - 6y + 9 = 0x
2
+y
2
−
4
25
x−6y+9=0
Comparing with general equation of circle:
x^2 + y^2 +2gx + 2fy + c = 0x
2
+y
2
+2gx+2fy+c=0
Which on solving yields-
g = \frac {-25}{8}
8
−25
, f = -3, c = 9
Radius, r = \sqrt {(\frac {25}{8})^2+9-9} = \frac {25}{8}
(
8
25
)
2
+9−9
=
8
25
Now right form of equation of normal :
y - y_1 = \frac {-y_1}{2a}(x-x_1)y−y
1
=
2a
−y
1
(x−x
1
)
⇒ (y-6) = \frac {-6}{\frac {9}{2}}(x-4)(y−6)=
2
9
−6
(x−4)
⇒ 3y-18 = -4x + 26
⇒ 4x + 3y -34 = 0 """ Equation of Normal
Therefore, Perpendicular distance of centre O (\frac {-25}{8}
8
−25
, 3) from normal = d
d = \left |\frac {4 \times \frac {25}{8}+3(3)-34}{\sqrt {4^2+32}}\right |
∣
∣
∣
∣
4
2
+32
4×
8
25
+3(3)−34
∣
∣
∣
∣
= \left |\frac {\frac {25}{2}-25}{5}\right |
∣
∣
∣
∣
5
2
25
−25
∣
∣
∣
∣
d = \frac {5}{2}
2
5
We know perpendicular from center to chord bisects it
So, AB = AC
In ΔABO, By Pythagoras theorem,
AO^2 = OB^2 + AB^2AO
2
=OB
2
+AB
2
(\frac {25}{8})^2 = (\frac {5}{2})^2 + AB^2 = \frac {15}{4}(
8
25
)
2
=(
2
5
)
2
+AB
2
=
4
15
here is your answer mate
Step-by-step explanation:
Answer is \frac {15}{4}
4
15
Step-by-step explanation:
Given, y^2 = 9xy
2
=9x
Comparing with standard equation of parabola : y^2 = 4axy
2
=4ax
4a = 9 or a = \frac {9}{4}
4
9
hence, Value for a = \frac {9}{4}a=
4
9
Therefore, Focus = s(\frac {9}{4}
4
9
, 0)
As we know the coordinates of end points of diameter, the equation of circle can be given by:
(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0(x−x
1
)(x−x
2
)+(y−y
1
)(y−y
2
)=0
⇒ (x-4)(x-\frac {9}{4}) + (y-6)(y-0) = 0(x−4)(x−
4
9
)+(y−6)(y−0)=0
⇒ (x-4)(4x-9) + 4y(y-6) = 0(x−4)(4x−9)+4y(y−6)=0
⇒ 4x^2 + 4y^2-25x-24y+36 = 04x
2
+4y
2
−25x−24y+36=0
⇒ x^2 + y^2 - \frac {25}{4}x - 6y + 9 = 0x
2
+y
2
−
4
25
x−6y+9=0
Comparing with general equation of circle:
x^2 + y^2 +2gx + 2fy + c = 0x
2
+y
2
+2gx+2fy+c=0
Which on solving yields-
g = \frac {-25}{8}
8
−25
, f = -3, c = 9
Radius, r = \sqrt {(\frac {25}{8})^2+9-9} = \frac {25}{8}
(
8
25
)
2
+9−9
=
8
25
Now right form of equation of normal :
y - y_1 = \frac {-y_1}{2a}(x-x_1)y−y
1
=
2a
−y
1
(x−x
1
)
⇒ (y-6) = \frac {-6}{\frac {9}{2}}(x-4)(y−6)=
2
9
−6
(x−4)
⇒ 3y-18 = -4x + 26
⇒ 4x + 3y -34 = 0 """ Equation of Normal
Therefore, Perpendicular distance of centre O (\frac {-25}{8}
8
−25
, 3) from normal = d
d = \left |\frac {4 \times \frac {25}{8}+3(3)-34}{\sqrt {4^2+32}}\right |
∣
∣
∣
∣
4
2
+32
4×
8
25
+3(3)−34
∣
∣
∣
∣
= \left |\frac {\frac {25}{2}-25}{5}\right |
∣
∣
∣
∣
5
2
25
−25
∣
∣
∣
∣
d = \frac {5}{2}
2
5
We know perpendicular from center to chord bisects it
So, AB = AC
In ΔABO, By Pythagoras theorem,
AO^2 = OB^2 + AB^2AO
2
=OB
2
+AB
2
(\frac {25}{8})^2 = (\frac {5}{2})^2 + AB^2 = \frac {15}{4}(
8
25
)
2
=(
2
5
)
2
+AB
2
=
4
15