Parabola MCQ Quiz - Objective Question with Answer for Parabola - Download Free PDF

Calculation:

Given the parabola

y2 = 4x 

and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is 

A standard parametric form for y2 = 4x is 

since  

The slope of the tangent at 

Differentiate y2 = 4x 

At the point  one has 

We require this slope to equal 1.

Now point of contact

Substitute t = 1 in . 

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.

Concept:

Focal Chord of Parabola and Circle Properties:

• The focal chord of the parabola passes through points on the parabola making a certain angle with the x-axis.
• The circle with one diameter as the focal chord touches the y-axis at a certain point.

• The tangent of the given angle and the coordinates of the point P on the parabola can be used to find the parameter t.
• The equation of the circle with diameter PS and its intersection with the y-axis is used to find

Calculation:

Given,

The parabola:

Focal chord PQ makes an angle of with positive x-axis.

Using slope

Coordinates of P:

Circle with diameter PS and focus S:

At x = 0,

Calculate

∴ The correct answer is Option 1.

Calculation

For the parabola , the equation of the tangent in slope form is:

where m is the slope of the tangent.

The tangent must also touch the circle

The perpendicular distance of (0, 0) to the tangent  should be equal to the radius of the circle √2.

The perpendicular distance d from the origin (0, 0) to the line is given by:

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Let 
⇒ 

⇒ (z + 2)(z - 1) = 0.

Thus  z = 1  (since z = -2 is not valid for )

⇒ 

Equation of the common tangents

For m = 1, the tangent equation is:

⇒ y = x + 2.

For m = -1, the tangent equation is:

⇒ y = -x + 2.

These are the equations of the two direct common tangents.

Points of tangency on the circle

For y = x + 2:

Substitute y = x + 2 into

⇒ 

⇒ 

⇒ 

⇒ 

Substituting x = -1 into y = x + 2

⇒ y = -1 + 2 = 1

So, the point of tangency is (-1, 1).

For y = -x + 2

Substitute y = -x + 2 into 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Substituting x = -1 into y = -x + 2

⇒ y = -(-1) + 2 = -1

So, the point of tangency is (-1, -1).

Thus, the points of tangency on the circle are P(-1, 1) and Q(-1, -1)

Points of tangency on the parabola

Substitute the tangent equations into the parabola y^2 = 8x to find the points of tangency:

For y = x + 2:

Substitute y = x + 2 into
   
⇒ 

⇒ 

⇒ 

⇒ 

Substituting x = 2 into y = x + 2

⇒ y = 2 + 2 = 4

So, the point of tangency is R(2, 4).

For y = -x + 2

Substitute y = -x + 2 into 

⇒ 

⇒ 

⇒ 

⇒ 

So, the point of tangency is S(2, -4)

The vertices of the quadrilateral are

P(-1, 1), Q(-1, -1), R(2, 4), S(2, -4)

PR =  = 3√2 

Hence Option(2) is the correct answer.

Calculation

Let the equation chord be

y² = 6x

⇒ 

Now as mentioned in the question, this line will pass from focus.

Substitute this in equation we get  ...(1)

Now Distance of this line from vertex is given to be so using the formula of perpendicular distance of a line from a point, we get

Taking square

 ...(2)

Solving both the equation we get

Hence option 1 is correct

Calculation

For the parabola , the equation of the tangent in slope form is:

where m is the slope of the tangent.

The tangent must also touch the circle

The perpendicular distance of (0, 0) to the tangent  should be equal to the radius of the circle √2.

The perpendicular distance d from the origin (0, 0) to the line is given by:

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Let 
⇒ 

⇒ (z + 2)(z - 1) = 0.

Thus  z = 1  (since z = -2 is not valid for )

⇒ 

Equation of the common tangents

For m = 1, the tangent equation is:

⇒ y = x + 2.

For m = -1, the tangent equation is:

⇒ y = -x + 2.

These are the equations of the two direct common tangents.

Points of tangency on the circle

For y = x + 2:

Substitute y = x + 2 into

⇒ 

⇒ 

⇒ 

⇒ 

Substituting x = -1 into y = x + 2

⇒ y = -1 + 2 = 1

So, the point of tangency is (-1, 1).

For y = -x + 2

Substitute y = -x + 2 into 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Substituting x = -1 into y = -x + 2

⇒ y = -(-1) + 2 = -1

So, the point of tangency is (-1, -1).

Thus, the points of tangency on the circle are P(-1, 1) and Q(-1, -1)

Points of tangency on the parabola

Substitute the tangent equations into the parabola y^2 = 8x to find the points of tangency:

For y = x + 2:

Substitute y = x + 2 into
   
⇒ 

⇒ 

⇒ 

⇒ 

Substituting x = 2 into y = x + 2

⇒ y = 2 + 2 = 4

So, the point of tangency is R(2, 4).

For y = -x + 2

Substitute y = -x + 2 into 

⇒ 

⇒ 

⇒ 

⇒ 

So, the point of tangency is S(2, -4)

The vertices of the quadrilateral are

P(-1, 1), Q(-1, -1), R(2, 4), S(2, -4)

PR =  = 3√2 

Hence Option(2) is the correct answer.

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:

The coordinates of the point where the chord cut the parabola satisfIes  the equation of a parabola.

Calculation:

Given:

The equation of a parabola is y² = x.

The angle made by Chord OA with x-axis is θ

Let the length of the chord OA of the parabola is L 

So, Length of AM = L sinθ 

and Length of OM = L cosθ 

So, The coordinate of A = (L cos θ, L sin θ)

And this point will satisfy the equation of parabola y2 = x.

⇒ (Lsin θ)² = L cos θ

⇒L² sin² θ = L cos θ

⇒ L = cos θ. cosec2 θ

∴ The required length of chord is cos θ. cosec2 θ.

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)

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:

The properties of a rectangular hyperbola  are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: 
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: 

CALCULATION:

Given: Equation of hyperbola is x2 - y2 =  1

As we can see that, the given hyperbola is a horizontal hyperbola.

So, by comparing the given equation of hyperbola with  we get

⇒ a = 1 and b = 1

As we know that, length of latus rectum of a hyperbola is given by 

So, the length of latus rectum of given hyperbola is 2 units.

Hence, option D is the correct answer.

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:

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:

The length of the latus rectum of the parabola y2 = 4ax is 4a.

Calculations:

Given, the parabola y2 = 4kx passes through point (-2, 1),

⇒The  point (-2, 1) is satisfying the equation of parabola y2 = 4kx 

⇒ (1)² = 4k (-2)

⇒ k = 

Now, the length of the latus rectum = 4k

⇒The length of latus rectum = 4()

⇒The length of latus rectum = 

The length of latus rectum can not be negative.

⇒The length of latus rectum = 

Hence, if parabola y2 = 4kx passes through the point (-2, 1), then the length of the latus rectum is .

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:

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