Latus Rectum MCQ Quiz in বাংলা - Objective Question with Answer for Latus Rectum - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 19, 2025
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
Answer (Detailed Solution Below)
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 ?
Answer (Detailed Solution Below)
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 ?
Answer (Detailed Solution Below)
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 represents a
Answer (Detailed Solution Below)
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 .
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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:
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.
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 ________.
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
Latus Rectum Question 9 Detailed Solution
\(y (t_{1} + t_{2}) - 2x - 2at_{1}t_{2} = 0\)
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 \(\rm x = \frac{y^2}{21}\) ?
Answer (Detailed Solution Below)
Latus Rectum Question 10 Detailed Solution
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.