Ellipse — Chapter Test | Class 11 IMO

Q1

The foci of x²/36 + y²/20 = 1 are:

c² = 36 − 20 = 16, so c = 4.

Q2

Find the equations of the directrices for x²/16 + y²/9 = 1.

a=4, e=√7/4. Directrices are x = ±a/e = ±4 / (√7/4) = ±16/√7.

Q3

The foci of x²/16 + y²/64 = 1 are:

c² = 64 − 16 = 48, so c = 4√3.

Q4

A shop owner designs an elliptical logo with a = 13 and b = 5. The focal distance from centre is:

c = √(169 − 25) = 12.

Q5

The foci of x²/50 + y²/18 = 1 are:

√32 = 4√2, so both forms are correct.

Q6

The focus lies inside the ellipse because:

Since e < 1, c < a.

Q7

Find the foci of 25x² + 9y² = 225.

x²/9 + y²/25 = 1. Vertical ellipse. c = √(25-9) = 4. Foci are (0, ±4).

Q8

If a point P(x₁, y₁) lies strictly inside the ellipse x²/a² + y²/b² = 1, then the value of x₁²/a² + y₁²/b² - 1 is:

For interior points, the expression S₁ = x₁²/a² + y₁²/b² - 1 is negative (< 0).

Q9

A bridge over a canal in Punjab is a semi-elliptical arch with a span of 20 m and maximum height of 5 m. What is the height of the arch at 4 m from the center?

x²/100 + y²/25 = 1. If x=4, 16/100 + y²/25 = 1 => y²/25 = 84/100 => y² = 21. Height = √21 m.

Q10

As an ellipse becomes flatter, its eccentricity approaches:

A circle has e=0. As it flattens into a line segment, c approaches a, so e = c/a approaches 1.

Q11

If an ellipse has e = 0.8 and a = 5, then c is:

c = ea = 4.

Q12

For x²/196 + y²/100 = 1, c equals:

c² = 96, c = 4√6 ≈ 9.8, nearest exact option is 10.

Q13

The line y = mx + c touches the ellipse x²/a² + y²/b² = 1 if:

Standard condition for tangency to an ellipse is c = ±√(a²m² + b²), hence c² = a²m² + b².

Q14

The distance between the foci of x²/25 + y²/16 = 1 is:

c = 3, distance = 2c = 6.

Q15

Determine the foci of x²/81 + y²/45 = 1.

a²=81, b²=45. c = √(81-45) = 6. Foci at (±6, 0).