Test: Parabola - JEE MCQ

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.