Test: Parabola (20 July) - 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 x² = 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.

Given directrix of parabola ⇒ x+y=2
and force is origin vertex is A(0,0)
We know that perpendicular distance from vertex  of the parabola to directrix is equal to 'a'  where 4a is the Latus Rectum of the 
Parabola 

x² − 2 = −2cost
⇒ x² = 2 − 2cost
⇒x² = 2(1−cost)
⇒x² = 2(1−(1−2sin² t/2))
⇒x² = 4sin² t/2
We have y = 4cos² t/2
⇒cos² t/2= y/4
We know the identity, sin² t/2 + cos² t/2 = 1
⇒ x²/4 + y/4 = 1
⇒ x² = 4−y represents a parabolic profile.

Given (t² , 2t) be one end of focal chord then other end be (1/t² , −2t)

Length of focal chord = [(t² - 1/t²)² + (2t + t/2)²]^½

= (t + 1/t)²

Focus of parabola y² = 8x is (2,0). Equation of circle with centre (2,0) is (x−2)² + y² = r²
Let AB is common chord and Q is mid point i.e. (1,0)
AQ² = y² = 8x
= 8×1 = 8
∴ r² = AQ² + QS²
= 8 + 1 = 9
So required circle is (x−2)² + y² = 9

Let E₁ = P₁ win the tournament, E² = P₂ reaches the semifinal since all players are equally skilled and there are 4 persons in the semifinal 
 both are in semifinal and P₁ wins in semifinal and final 

Eliminating t, we get (3x - 4y + 2)² = 16x + 12y - 27. Vertex is 
Hence  k = 29 and the area is 3π.

Taking the new axes as  we see that the parabola can be  with the required condition |X| <  200

Center of such circle in the case of parabolas  y² = 4ax, x² = 4ay is 

where d is the distance between lines whose equations are ax+by+C₁ = 0 & ax + by + C₂ ​= 0

= 5
d = 5
If the distance between vertex and latus rectum = distance of vertex from directrix = a then d = 2a = 5
⇒ Length of latus rectum = 4a = 10