Test: Parabola - Grade 12 MCQ

CONCEPT:

The following are the properties of a parabola of the form: y² = - 4ax where a > 0

• Focus is given by (- a, 0)
• Vertex is given by (0, 0)
• Equation of directrix is given by: x = a
• Equation of axis is given by: y = 0
• Length of latus rectum is given by: 4a
• Equation of latus rectum is given by: x = - a

CALCULATION:
Given: Equation of parabola is y² = - 12x
The given equation can be re-written as: y² = - 4 ⋅ 3 ⋅ x---------(1)
Now by comparing the equation (1), with y² = - 4ax we get
⇒ a = 3
As we know that, the length of latus rectum of a parabola is given by: 4a
So, length of latus rectum of the given parabola is: 4 ⋅ 3 = 12 units
Hence, option B is the correct answer.

Concept:
For a parabola symmetric about y-axis 
(y - k)2 = 4a (x - h)
Where vertex is (h, k) and focus is (h + a, k)
Calculation:
Given vertex is (3, 0)
h = 3, k = 0 
For focus = (5, 0)
h + a = 5
⇒ 3 + a = 5 
⇒ a = 2
So equation of parabola 

y² = 8x - 24

Concept:
Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Calculation:
Given: y² = -12x
⇒ y²= 4 × (-3) × x
Compare with standard equation of parabola y² = 4ax
So, a = -3
Therefore, Focus  = (a, 0) = (-3, 0)

Concept:
Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Calculation:
Given: x² = 20y
⇒ x² = 4 × 5 × y
Compare with standard equation of parabola x² = 4ay 
So, 4a = 4 × 5
Therefore, Length of Latus rectum = 4a = 4 × 5 = 20

Concept:

Calculation:
Comparing the given equation (y - 3)² = 20(x - 1) with the general equation of the parabola (y - k)² = 4a(x - h), we can say that:
k = 3, a = 5, h = 1.
Vertex is (h, k) = (1, 3).

Concept:

Latus rectum:
The latus rectum of a conic section is the chord (line segment) that passes through the focus, is perpendicular to the major axis, and has both endpoints on the curve.

• Length of Latus Rectum of Parabola y² = 4ax is 4a
• End points of the latus rectum of a parabola are L = (a, 2a), and L’ = (a, -2a)

Calculation:
Given equation:
y² − 8x + 6y + 1 = 0
⇒ y² + 6y + 9 - 9 - 8x + 1 = 0
⇒ (y + 3)² - 8x - 8 = 0
⇒ (y + 3)² = 8x + 8
⇒ (y + 3)² = 8 (x + 1)
Let new coordinate axes be X and Y,
Here X = x + 1 and Y = y + 3
⇒ Y² = 4aX
Now comparing with above equation,
∴ 4a = 8 ⇒ a = 2
Focus: (a, 0)
X = a  and Y = 0
⇒ x + 1 = 2 and y + 3 = 0
⇒ x = 1 and y = -3
∴ focus of parabola is (1, -3)

Concept:
Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Concept:
Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Calculation:
Given: x² = 16y
⇒ x² = 4 × 4 × y
Compare with standard equation of parabola x² = 4ay 
So, a = 4
Therefore, Focus  = (0, a) = (0, 4)

Comparing equation with y²= -4ax.
4a = 100 => a = 25.
Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is x = a => x = 25.

Comparing equation with y² = 4ax.
4a = 100 => a = 25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is x = -a => x = -25.