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Conic Sections - Latus rectum (Parabola)

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, passing through the focus and whose endpoints lie on the parabola.

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For the standard parabola y2=4axy^2 = 4ax, the axis of symmetry is the xx-axis and the focus is at (a,0)(a, 0).

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The length of the latus rectum is the absolute value of the coefficient of the linear variable in the standard equation (e.g., in y2=4axy^2 = 4ax, it is ∣4a∣|4a|).

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The latus rectum is bisected by the focus. For y2=4axy^2 = 4ax, the segment extends 2a2a units above and 2a2a units below the focus.

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The semi-latus rectum is half the length of the latus rectum, which is 2a2a for the parabola y2=4axy^2 = 4ax.

📐Formulae

Length of Latus Rectum for y2=4ax is 4a\text{Length of Latus Rectum for } y^2 = 4ax \text{ is } 4a

Length of Latus Rectum for x2=4ay is 4a\text{Length of Latus Rectum for } x^2 = 4ay \text{ is } 4a

Endpoints of Latus Rectum for y2=4ax:(a,2a) and (a,−2a)\text{Endpoints of Latus Rectum for } y^2 = 4ax: (a, 2a) \text{ and } (a, -2a)

Endpoints of Latus Rectum for x2=4ay:(2a,a) and (−2a,a)\text{Endpoints of Latus Rectum for } x^2 = 4ay: (2a, a) \text{ and } (-2a, a)

💡Examples

Problem 1:

Find the length of the latus rectum and the coordinates of its endpoints for the parabola y2=12xy^2 = 12x.

Solution:

Comparing y2=12xy^2 = 12x with y2=4axy^2 = 4ax, we get: 4a=124a = 12 a=124=3a = \frac{12}{4} = 3 Length of latus rectum =4a=12= 4a = 12. Focus is at (a,0)=(3,0)(a, 0) = (3, 0). Endpoints are (a,±2a)=(3,±2×3)=(3,±6)(a, \pm 2a) = (3, \pm 2 \times 3) = (3, \pm 6). Thus, the endpoints are (3,6)(3, 6) and (3,−6)(3, -6).

Explanation:

First, identify the value of aa by comparing the given equation to the standard form. The coefficient of xx directly gives the length of the latus rectum. The coordinates are then derived using the focus as the midpoint.

Problem 2:

Find the length of the latus rectum for the parabola x2=−16yx^2 = -16y.

Solution:

The equation is of the form x2=−4ayx^2 = -4ay. Comparing coefficients: 4a=164a = 16 a=4a = 4 Length of latus rectum =4a=16= 4a = 16.

Explanation:

In the form x2=−4ayx^2 = -4ay, the length of the latus rectum is always the positive value ∣4a∣|4a|. Here, 4a=164a = 16, so the length is 1616 units.

Problem 3:

If the length of the latus rectum of a parabola y2=4axy^2 = 4ax is 88, find the coordinates of the focus.

Solution:

Given length of latus rectum =8= 8. 4a=84a = 8 Dividing by 44: 8÷42\begin{array}{r} 8 \\ \div 4 \\ \hline 2 \end{array} So, a=2a = 2. The focus for y2=4axy^2 = 4ax is (a,0)(a, 0). Therefore, focus =(2,0)= (2, 0).

Explanation:

The length of the latus rectum is 4a4a. By solving for aa, we find the distance from the vertex to the focus, which gives us the focus coordinates for a standard parabola.