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

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.

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

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.

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).

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. 

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

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

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:

When the equation of a parabola is given by y² = 4ax......(equation 1.)

Then Latus rectum is, L = 4a.

Calculation:

The given equation of a parabola is y² + 4x + 4y + k = 0.

It can be written as

y² + 4y + 4x + k = 0

⇒ y² + (2 × 2y) + 4x + k = 0

⇒ y² + 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)² = - 4x - k + 4

⇒ (y + 2)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.

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

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

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),

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

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

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

Concept:

Equation of parabola having a vertex at the origin and along X-axis: y2 = 4ax,

Length of latus rectum = 4a

Calculation:

Standard form of parabola: y2 = 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.