Test: Equation of Parabola - Grade 12 MCQ

Concept:
Equation of normal to a curve at point (h, k) on the curve is

Equation of tangent to a curve at point (h, k) on the curve is

Calculation:

Given equation of the parabola y² = 12x  ...(i)

Differentiate both sides with respect to x, 

Equation of normal to a curve at point (h, k) on the curve is

∴ Equation of normal to the parabola at point A( 3, - 6) on the curve is

⇒ y + 6 = x - 3
⇒ x = y + 9
Substituting x in eq (i)
⇒ y² = 12 (y + 9)
⇒ y² - 12y - 108 = 0
⇒ y² - 18 y + 6 y - 108 = 0
⇒ y( y - 18) + 6 (y - 18) = 0
⇒ (y + 6) ( y - 18) = 0
⇒ y = 18, - 6
We take y = 18, substitute y in eq(i)
18² = 12x 
⇒ x = 27
Hence point P where the normal cuts parabola again is (27, 18).
Equation of tangent to a curve at point (h, k) is

We know from eq(ii) that dy/dx = 6/y


The correct answer is option (A).

The equation of tangent to the parabola x² = 4ay can be written as y = mx - am² 
For the Parabola y = x², we can written of the equation of the tangent as y = mx - m²/4
Similarly, for the Parabola y = -(x - 2)², we can written of the equation of the tangent as 

As the tangent is common to both the parabola, both the equations should represent the same line.
Hence,

⇒ m(m - 4) = 0 
⇒ m = 0, 4
Hence, the common tangent are y = 0 and y = 4x - 4
⇒ y = 4(x - 1)
∴ The correct option is (d)

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

According to the data given above, the equation of the parabola will be y² = 16x
Here, 4a = 16
⇒ a = 4
So, The equation of directrix of the parabola is x = -a
⇒ x = -4
∴ The correct option is (d)

According to the data given above, the equation of the parabola will be y² = 16x
Given that, inclination of the tangent is θ = 60^o
⇒ m = tan60^o = √3
∴ Equation of the tangent is a y = mx + a/m
Here, a = 4 and m = √3

∴ The correct option is (a)

Concept:

• Focus is the mid point of the latus rectum L and L'
• Length of the latus rectum = 4a

Given:
The end points of the latus rectum of L(4, 8) and L'(4, -8)
Focus is the mid point of the latus rectum L(4, 8) and L'(4, -8)

Which is the point on the x-axis.
∴ The axis of the parabola is x-axis.
Now, Length of the latus rectum = 8 + 8 = 16 ⇒ 4a = 16
⇒ a = 4

The distance of the focus from the vertex = 4
∴ Vertex is (0, 0)
By the data given, the parabola is open right.
∴ The equation of the parabola with vertex (0, 0) axis along x-axis and open right is y² = 4ax
⇒ y² = 16x
Which is the equation of the required parabola.
∴ The correct option is (2)

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
(y - k)² = 4a(x - h)
⇒ (y - 0)² = 4 x 2(x - 3) 
⇒ 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: 
Since the vertex is at (0, 0) and the focus is at (0, 4) which lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, equation of the parabola is of the form x² = 4ay.
x² = 4 × 4y
∴ x² = 16y.

Concept:

Equation of parabola along Y-axis: (x – h)² = ±4a (y - k),

Here, point (h, k) is vertex of parabola

4a = length of latus rectum and focus (0, ±a)

Calculation:

Here, focus = (0, 2)

So a = 2

(h, k) = (0, 3)

As the parabola facing downward along Y-axis, equation will be

(x – h)² = -4a (y - k)

Here, a = Distance of vertex form focus = 1

⇒ (x – 0)² = -4 (1)(y - 3)

⇒ x² = -4y + 12

⇒ x² + 4y = 12

Hence, option (c) is correct.

Concept used:

For a parabola symmetric about the y-axis 

(y − k)² = 4a(x − h)

Where vertex is (h, k) and focus is (h + a, k)

Calculation:

Given vertex is (4, 0)

h = 4, k = 0 

For focus = (6, 0)

⇒ h + a = 6

⇒ 4 + a = 6 

⇒ a = 2

So equation of parabola 

(y − k)²  = 4a(x  −  h)

⇒ (y − 0)²  = (4 × 2)(x − 4)

⇒ y² = 8x − 32

Concept:

For a parabola symmetric about x-axis 

(y − k)² = 4a(x − h)

Where,

Focus is (h + a, k)

Vertex is (h, k) 

Calculation:

Given:

Focus = (4, 0)

Vertex = (2, 0)

Given vertex is (2, 0)

h = 2, k = 0 

For focus = (4, 0)

h + a = 4

⇒ 2 + a = 4 

⇒ a = 2

So equation of parabola 

(y − k)² = 4a(x − h)

⇒ (y − 0)² = 4 × 2(x − 2)

⇒ y² = 8x − 16

∴ The equation of parabola is y² = 8x – 16.