Conic Sections Test - 3

According to definition of parabola , is is the locks of the points in that planes that are equidistant from both focus and directrix.

Given, focus : (-3,0)
directrix : x + 5 = 0
Let (x ,y) is the point on the parabola .
∴distance of point from focus = distance of point from directrix
⇒√{(x + 3)²  + y² } = |x + 5|/√(1²  + 0² )
⇒√{(x + 3)² + y² } = |x + 5|
squaring both sides,
(x + 3)² + y² = (x + 5)²
⇒y² = (x + 5)² - (x + 3)²
⇒y² = (x + 5 - x - 3)(x + 5 + x + 3)
⇒y² = 2(2x + 8) = 4(x + 4)

Hence, equation of parabola is y² = 4(x + 4)

As the quadrants lies in first and second quadrant
y = -a   Focus(0,a)
x^{2 = 4ay}
 

y² +3=2(2x+y) represents parabola.
y² +3=4x+2y
y² −2y+3=4x
y² −2y+1+3=4x+1
(y −1)² =4x −2
(y −1)² =4(x −1/2)
So, the vertex of parabola=(1/2,1) and axis is parallel to x axis.
a=1
Focus=(1/2+1,1)
=(3/2,1)

Given the vertex of the parabola is (0,0) and focus is at (0,-2).
This gives the axis of the parabola is the positive y −axis.
Then the equation of the parabola will be x^2 = 4ay where a = -2.
So the equation of the parabola is x2 = -8y.

The chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.

When a is positive, the parabola is opening to the right and since directrix is x=-a, it gives the answer.

x+y=3 m=−1
(−2,1)m = 1
y −1 = 1(x+2)
y −1 = x+2
x −y+3 = 0
x −y+3 = 0
x+y −3 = 0
2x = 0
x = 0; y = 3
Vertex=
Midpoint of focus and directrix = (0 −2)/2,(3+1)/2
​=(−1,2)

f(0,1),d(x+2=0)
Distance of any point on parabola and focus is equal to distance of point and directrix.
fP=(h −0)² +(k −1)² = (h² +k² +1 −2k)^1/2
Distance of point (h,k) and line x+2=0
Using point line distance formula.
dP=h+2
[h² +k² +1 −2k]^1/2 =h+2
h² +k² +1 −2k = h² +4+4h
k² −2k+1 −4 −4h=0
replacing h →x,k →y  y² −2y+1 −4 −4x=0
(y −1)² =4(x+1)   …(1)
Let Y=y −1,X=x+1 then (1) becomes  
Y^2=4aX²
Here a=1 any point on this parabola will be of the form (at² ,2at)=(t² ,2at)
⇒X=t² ⇒x+1=t²
⇒x=t² −1
⇒Y² =2t
⇒y −1 = 2t ⇒y = 2t+1
∴Any point on the parabola (y −1)² =4(x+1) is  
= (t² −1,2t+1)

Comparing the equation with the standard form ax² +2hxy+by² +2gx+2fy+c=0
a=2,h=−3/2,b=5,g=3,f=−3/2,c=5
Δ=abc+2fgh −af² −bg² −ch²
=(2)(5)(5)+2(−3/2)(3)(−3/2)−(2)(−3/2)² −(5)(3)² −(5)(−3/2)²
=50+27/2 −9/2 −45 −225/4
=−169/4 is not equal to 0
Descriminant =h² −ab
=(−3/2)² −(2)(5)
= 9/4 −10
= −31/4 <0
So, the curve represents either a circle or an ellipse
a is not equal to b and  
Δ/a+b = −(169/4)/2+5
=−169/28 <0
So, the curve represents a ellipse.