Latus Rectum MCQ Quiz in বাংলা - Objective Question with Answer for Latus Rectum - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Mar 19, 2025

পাওয়া Latus Rectum उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Latus Rectum MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Latus Rectum MCQ Objective Questions

Top Latus Rectum MCQ Objective Questions

Latus Rectum Question 1:

The equation of latus rectum of a parabola is x + y = 8 and the equation of the tangent at the vertex is x + y = 12, then length of the latus rectum is

  1. 4√2
  2. 2√2
  3. 8
  4. 8√2

Answer (Detailed Solution Below)

Option 4 : 8√2

Latus Rectum Question 1 Detailed Solution

Concept:

We know the equation of the latus rectum of a parabola passes through the focus of the parabola.

The equation of the latus rectum and the equation of the tangent of the vertex are parallel and the distance between these two lines gives focal length.

And the formula of the distance between the two parallel lines ax + by = c1 and ax + by = c2 is given by

→ \(\frac {|c_{2}-c_{1}|}{√{a^2+b^2}}\) .......(equation 1)

Calculation:

Given:

Equation of Latus rectum of parabola: x + y = 8.

Equation of the tangent at the vertex: x + y = 12.

The length of the latus rectum of the parabola = 4 × Distance between the latus-rectum and the tangent at the vertex.

We can see that equation of latus rectum and the equation of vertex are parallel.

Thus, the distance between the latus rectum and the tangent at the vertex is = \(\frac {12-8}{√{1^2+1^2}}\)  

= 2√2  ---(Using equation 1)

So, the length of the latus rectum is = 4 × 2√2  = 8√2.

Latus Rectum Question 2:

Find the equation of the latus rectum of the parabola y2 = - 6x ?

  1. x = 3/2
  2. x = - 3/2
  3. y = 3/2
  4. y = - 3/2

Answer (Detailed Solution Below)

Option 2 : x = - 3/2

Latus Rectum Question 2 Detailed Solution

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 the parabola is y2 = - 6x

The given equation of parabola can be re-written as: y2 = - 4 ⋅ (3/2)x----------(1)

Now by comparing the equation (1) with y2 = - 4ax we get

⇒ a = 3/2

As we know that, equation of latus rectum of the parabola of the form y2 = - 4ax is given by: x =  - a

So, equation of latus rectum of given parabola is: x = - 3/2

Latus Rectum Question 3:

Find the length of the latus rectum of the parabola 5y2 = - 16x ?

  1. 13/5
  2. 19/5
  3. 16/5
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 16/5

Latus Rectum Question 3 Detailed Solution

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 5y2 = - 16x

The given equation can be re-written as: y2 = - 4 ⋅ (4/5) ⋅ x---------(1)

Now by comparing the equation (1), with y2 = - 4ax we get

⇒ a = 4/5

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 ⋅ (4/5) = 16/5 units

Hence, option C is the correct answer.

Latus Rectum Question 4:

Comprehension:

Parametric equation is given as (3 + t2, 2t - 1).

Parametric equation represents a 

  1. Parabola with focus at (3, -1)
  2. Parabola with vertex at (3, -1)
  3. Ellipse with centre at (3, -1)
  4. None

Answer (Detailed Solution Below)

Option 2 : Parabola with vertex at (3, -1)

Latus Rectum Question 4 Detailed Solution

Concept:

If the equation of parabola is in the form of (y – k)2 = 4a(x – h) then

Endpoints of lactus rectum = (h + a, k ± 2a), 

Focus of parabola = (h + a, k)

Calculation:

Given:

x = 3 + t2, and y = 2t - 1

t2 = x - 3 and t = \(\frac{y+1}{2}\)

Eliminating t, we get,

(y + 1)2 = 4(x - 3)

Which is a parabola with vertex at (3, -1).

∴ The parametric equation represents a  parabola with vertex at (3, -1).

Latus Rectum Question 5:

Find the length of latus rectum of parabola y2 = - 24x . 

  1. 16 units
  2. 12 units
  3. 6 units
  4. 24 units

Answer (Detailed Solution Below)

Option 4 : 24 units

Latus Rectum Question 5 Detailed Solution

Concept: 

Parabola, y2 = 4ax, where a > 0, then  

Length of latus rectum = 4a 

Parabola, y2 = - 4ax, where a > 0, then 

Length of latus rectum = 4a 

Calculation: 

Given parabolic equation , y2 = - 24x 

On comparing with standard equation , a = 6 

We know that , length of latus rectum = 4a

Latus rectum = 4a 

⇒ Latus rectum = 4 × 6  

⇒ Latus rectum = 24 units.

The correct option is 4. 

Latus Rectum Question 6:

The equations of the normals at the ends of the latus-rectum of the parabola \(y^2=4ax\) are given by

  1. \(x^2-y^2-6ax+9a^2=0\)
  2. \(x^2-y^2-6ax-6ay+9a^2=0\)
  3. \(x^2-y^2-6ay+9a^2=0\)
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : \(x^2-y^2-6ax+9a^2=0\)

Latus Rectum Question 6 Detailed Solution

The coordinates of the ends of the latus-rectum of the parabola \(y^2=4ax\) are \((a, 2a)\) respectively. The equation of the normal at \((a, 2a)\) to \(y^2=4ax\) is

\(y-2a=\dfrac{-2a}{2a}(x-a)\)

\(x+y-3a=0\)

Similarly, the equation of the normal \((a, -2a)\) is

\(x-y-3a=0\)

Then combined equation is

\(x^2-y^2-6ax+9a^2=0\)

Latus Rectum Question 7:

The equation y2 + 4x + 4y + k = 0 represents a parabola whose lauts rectum is

  1. 1
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 4 : 4

Latus Rectum Question 7 Detailed Solution

Concept:

  • A latus rectum is a straight line that passes through the focus of the parabola and is perpendicular to the axis of the parabola.

It can be seen in the following figure:

openleftwardparabola

 

When the equation of a parabola is given by y= 4ax......(equation 1.)

Then Latus rectum is, L = 4a.

Calculation:

The given equation of a parabola is y+ 4x + 4y + k = 0.

It can be written as

y+ 4y + 4x + k = 0

⇒ y+ (2 × 2y) + 4x + k = 0

⇒ y+ 2 × 2y + 4 - 4 + 4x + k = 0

(Adding and subtracting 4 in the left hand side of the equation)

⇒ (y2 + 2×2y + 4) - 4 + 4x + k = 0

⇒ (y + 2)= - 4x - k + 4

⇒ (y + 2)= - 4 (x - \(\frac{4+k}{4}\))

On comparing equation 1, we get a = -1

⇒ Latus rectum = 4a = -4

and Taking magnitude, Latus rectum = 4 units.

Latus Rectum Question 8:

If the lengths of the transverse axis and the latus rectum of a hyperbola are \(6\) and \(\dfrac{8}{3}\) respectively, then the equation of the hyperbola is ________.

  1. \(4x^2-9y^2=72\)
  2. \(4x^2-9y^2=36\)
  3. \(9x^2-4y^2=72\)
  4. \(9x^2-4y^2=36\)

Answer (Detailed Solution Below)

Option 2 : \(4x^2-9y^2=36\)

Latus Rectum Question 8 Detailed Solution

Let the equation of the given hyperbola be \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

Then the lengths of its latus rectum and its transverse axis are \(\frac{2b^2}{a}\) and \(2a\) respectively.

So, we have \(2a = 6\). That is, \(a = 3\).

Also, \(\frac{2b^2}{a} = \frac{8}{3}\)

i.e, \(b^2 = \frac{4a}{3}\)

i.e, \(b^2 = 4\)

So, the equation of the hyperbola is \(\frac{x^2}{9}-\frac{y^2}{4} = 1\).

That is, \(4x^2 + 9y^2 = 36\).

Latus Rectum Question 9:

The intercept of the latus rectum to the parabola \(y^{2} = 4ax\) are \(b\) and \(k\), then \(k\) is equal to:

  1. \(\dfrac {ab}{a - b}\)
  2. \(\dfrac {a}{b - a}\)
  3. \(\dfrac {b}{b - a}\)
  4. \(\dfrac {ab}{b - a}\)

Answer (Detailed Solution Below)

Option 4 : \(\dfrac {ab}{b - a}\)

Latus Rectum Question 9 Detailed Solution

For latus rectum \(PQ\),

\(y (t_{1} + t_{2}) - 2x - 2at_{1}t_{2} = 0\)

and \(t_{1}t_{2} = -1\)

For any point \(P(x, y)\), the focal distance is \(a + x\).

\(\therefore b = a + x = a + at_{1}^{2} = a (1 + t_{1}^{2})\) ...... (i)

\(c = a + x = a + at_{2}^{2} = a + \dfrac {a}{t_{1}^{2}} (\because t_{1}t_{2} = -1)\)

\(= \dfrac {a(1 + t_{1}^{2})}{t_{1}^{2}}\) ..... (ii)

\(\therefore \dfrac {b}{c} = t_{1}^{2}\)

\(\therefore\) From Eq. (i),

\(b = a\left (1 + \dfrac {b}{c}\right )\)

\(\Rightarrow b = a + \dfrac {ab}{c}\)

\(\therefore c = \dfrac {ab}{b - a}\)

Latus Rectum Question 10:

What is the length of the latus rectum of the curve \(\rm x = \frac{y^2}{21}\) ?

  1. 84 units
  2. 21 units 
  3. \(\sqrt{21}\) units 
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 21 units 

Latus Rectum Question 10 Detailed Solution

Concept:

Equation of parabola having a vertex at the origin and along X-axis: y= 4ax,

Length of latus rectum = 4a

 

Calculation:

Standard form of parabola: y= 4ax          ....(1)

We have , \(\rm x = \frac{y^2}{21}\) \(\Rightarrow \rm y^2=21x\)

By comparing the given equation \(\rm y^2=21x\) with equation (1), we get,

4a = 21

∴The length of the latus rectum is 21 units.

Hence, option (2) is correct.

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