Solve it guys please.......❤❤❤❤❤❤❤❤❤
✅ •• find the equation of parabola if the vertex us at (2,1) and the directrix is x=y-1
✅ •• find the equation of parabola if the focus is (-6 ,-6) and (-2,2) is the vertex....
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Answers
Answer:
x² + y² - 14x + 2y + 2xy + 17 = 0
4x² + y² - 4xy + 104x + 148y - 124 = 0
Step-by-step explanation:
(1)
Given Equation is x = y - 1
It can be written as, x - y = -1 ---- (i)
-y = -x - 1
y = x + 1
∴ Slope = 1.
∴ Slope of axis = -1.
Equation of axis is y - 1 = -1(x - 2)
y - 1 = -x + 2
y - 1 + x - 2 = 0
x + y - 3 = 0 ---- (ii)
On solving (i) & (ii), we get
x - y = -1
x + y = 3
------------
2x = 2
x = 1
Substitute x = 1 in (i), we get
x - y = -1
1 - y = -1
y = 2
∴ The directrix and axis intersects at A(1,2)
Let h(s,k) be the focus of parabola.
∴ V(2,1) is the mid-point of AS.
∴ 2 = 1 + h/2 and 1 = 2 + k/2
h = 3, k = 0.
∴ Focus is S(3,0).
Let P(x,y) be any point on parabola.
PS = PM
⇒√(x - 3)² + (y - 0)² = |x -y + 1/√1 + 1|
On Squaring both sides, we get
⇒ (x - 3)² + y² = (x - y + 1)^2/2)
⇒ x² + 9 - 6y + y² = x² + y² + 1 - 2xy - 2y + 2x/2
⇒ 2x² + 18 - 12y + 2y² = x² + y² + 1 - 2xy - 2y + 2x
⇒ x² + y² - 14x + 2y + 2xy + 17 = 0
--------------------------------------------------------------------------------------------
(2)
Let the focus be S(-6,-6) and vertex be A(-2,2)
Let z(x,y) be the projection of S on directrix. A is the mid-point of SZ.
⇒ (-2,2) = (-6 + x/2, -6 + y/2)
⇒ x = 2, y = 10.
∴ Z = (2,10)
Slope of directrix = -1/(-6 - 2/-6 + 2)
= -1/2..
Equation of directrix is y - 10 = -1/2(x - 2)
⇒ 2y - 20 = -x + 2
⇒ 2y - 20 + x - 2 = 0
⇒ x + 2y - 22 = 0.
Let P(x,y) be any point on the parabola, then
SP = PM
SP² = PM²{PM is the perpendicular from P to the directrix}
⇒ (x + 6)² + (y + 6)² = (x+ 2y - 22)²/1² + 4
⇒ x² + 36 + 12x + y² + 36 + 12y = x² + 4xy - 44x + 4y² - 88y + 484/5
⇒ 5(x² + 36 + 12x + y² + 36 + 12y) = x² + 4xy - 44x + 4y² - 88y + 484
⇒ 5x² + 60x + 5y² + 60y + 360 = x² + 4xy - 44x + 4y² - 88y + 484
⇒ 5x² + 104x - 4yx + y² + 148y - 124 - x² = 0
⇒ 4x² + y² - 4xy + 104x + 148y - 124 = 0
(or)
(2x - y)² + 104x + 148y - 124 = 0
Hope it helps!
Step-by-step explanation:
Given Equation is x = y - 1
It can be written as, x - y = -1 ---- (i)
-y = -x - 1
y = x + 1
∴ Slope = 1.
∴ Slope of axis = -1.
Equation of axis is y - 1 = -1(x - 2)
y - 1 = -x + 2
y - 1 + x - 2 = 0
x + y - 3 = 0 ---- (ii)
On solving (i) & (ii), we get
x - y = -1
x + y = 3
------------
2x = 2
x = 1
Substitute x = 1 in (i), we get
x - y = -1
1 - y = -1
y = 2
∴ The directrix and axis intersects at A(1,2)
Let h(s,k) be the focus of parabola.
∴ V(2,1) is the mid-point of AS.
∴ 2 = 1 + h/2 and 1 = 2 + k/2
h = 3, k = 0.
∴ Focus is S(3,0).
Let P(x,y) be any point on parabola.
PS = PM
⇒√(x - 3)² + (y - 0)² = |x -y + 1/√1 + 1|
On Squaring both sides, we get
⇒ (x - 3)² + y² = (x - y + 1)^2/2)
⇒ x² + 9 - 6y + y² = x² + y² + 1 - 2xy - 2y + 2x/2
⇒ 2x² + 18 - 12y + 2y² = x² + y² + 1 - 2xy - 2y + 2x
⇒ x² + y² - 14x + 2y + 2xy + 17 = 0