Question 17 of 70beginner🔧 ApplyShort Answer2 marks

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: x² = 6y

Correct Answer

The given equation is x² = 6y.

Finding the value of a: Comparing with the standard form x² = 4ay, we get 4a = 6, therefore a = 3/2.

Coordinates of the focus: Since the parabola is of the form x² = 4ay with positive coefficient, it opens upwards. The focus is at (0, a) = (0, 3/2).

Axis of the parabola: The axis of the parabola is the y-axis.

Equation of the directrix: The directrix is y = -a = -3/2.

Length of the latus rectum: Length of latus rectum = 4a = 6.

Exercise: EXERCISE 10.2 | Q: 2 | (Page 11)
For More Understanding

Explanation

The question asks to find properties of the parabola x² = 6y. From the textbook context, Example 5 demonstrates the method for y² = 8x, comparing with standard form y² = 4ax to find a = 2. Similarly, for x² = 6y, we compare with x² = 4ay to get 4a = 6, hence a = 3/2. The context states that for a parabola, the latus rectum length equals 4a, and the focus lies at distance 'a' from the vertex along the axis of symmetry.

Solution Steps

  1. Step 1: Compare x² = 6y with standard form x² = 4ay to get 4a = 6

  2. Step 2: Solve for a = 6/4 = 3/2

  3. Step 3: Focus is at (0, a) = (0, 3/2) since parabola opens upward

  4. Step 4: Axis is y-axis (the axis of symmetry)

  5. Step 5: Directrix is y = -a = -3/2

  6. Step 6: Length of latus rectum = 4a = 6

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