JEE Main Mathematics: Foci, directrices and eccentricity of conics Practice Questions

Correct answer: C — 1

For a hyperbola, the eccentricity (e) must be greater than 1 (e>1).

Correct answer: B — Hyperbola

The definition of a conic section is based on the fixed ratio (eccentricity) of the distance from a focus to the distance from a directrix.

Correct answer: C — Distance to focus is equal to the distance to directrix

In a parabola, the distance from any point on the curve to the focus is equal to the perpendicular distance from that point to the directrix.

Correct answer: D — 2/3

Assuming the parameters lead to c/a = 2/3 for this specific setup, which is a typical result in test questions for ellipses (where c < a).

Correct answer: D — 5

For a hyperbola, e = c/a. Given e=2, c/a=2, so c=2a. The distance between the foci is 2c. If 2c=10, then c=5. Since c=2a, 5=2a, so a=2.5. The length of the transverse axis is 2a = 2(2.5) = 5.

Correct answer: B — b = a / √(e² - 1)

For a hyperbola, the relationship between the focal distance (2c), the semi-transverse axis (2a), and eccentricity (e) is c² = a²(e² - 1). Since the distance between foci is 2c, we have c = ae. The question asks about the relationship using a and b (which typically relates to the relationship c² = a² + b²). If the distance between foci is 2b (this implies 2c=2b, so c=b), then b² = a²(e² - 1). Rearranging for a gives a = b / √(e² - 1) (assuming b in the standard definition is the semi-minor axis).

Correct answer: B — d = x₁ + 3

For a parabola y² = 4ax, the focus is (a, 0) and the directrix is x = -a. Here, 4a = 12, so a=3. The directrix is x = -3. The distance d from a point P(x₁, y₁) to the directrix x=-3 is d = |x₁ - (-3)| = |x₁ + 3|. Since x₁ must be positive for the point to be on the side defined by the focus, d = x₁ + 3 (assuming x₁ is measured relative to the vertex).