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51.

Equation of the hyperbola with eccentricty 3/2 and foci at (± 2, 0) is(A) x2/4 -y2/5 = 4/9(B) x2/9 - y2/9 = 1 (C) x2/4 - y2/9 = 1(D) none of these

Answer»

Answer is (A)
Discussion
52.

The distance between the foci of a hyperbola is 16 and its eccentricity is √2 . Its equation is(A) x2 – y2 = 32 (B)x2/4 - y2/9 = 1(C) 2x – 3y2 = 7 (D) none of these

Answer»

Answer is (A)
Discussion
53.

Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3

Answer»

Focus = (0, –3); directrix y = 3
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x2 = 4ay or x2 = – 4ay.
It is also seen that the directrix, y = 3 is above the x-axis, while the focus (0, –3) is below the x-axis. Hence, the parabola is of the form x2 = –4ay.
Here, a = 3
Thus, the equation of the parabola is x2 = –12y.
Discussion
54.

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

Answer»

Vertex (0, 0) focus (–2, 0)
Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, xaxis is the axis of the parabola, while the equation of the parabola is of the form y2 = – 4ax.
Since the focus is (–2, 0), a = 2.
Thus, the equation of the parabola is y2 = –4(2)x, i.e., y2 = –8x
Discussion
55.

The equation of the parabola having (2, 4) and (2, -4) as end points of its latus rectum is ________ (A) y2 = 4x (B) y2 = 8x (C) y2 = -16x (D) x2 = 8y

Answer»

Correct option is: (B) y2 = 8x

The given points lie in the 1st and 4th quadrants. 

∴ Equation of the parabola is y2 = 4ax 

End points of latus rectum are (a, 2a) and (a, -2a) 

∴ a = 2 

∴ required equation of parabola is y2 = 8x
Discussion
56.

If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is A. 2/3 B. 4/3C. 1/3 D. 4

Answer»

y2 = 4ax

22 = 4a (3)

4 = 12a

a = 4/12 = 1/3

Length of Latus Rectum = 4a = 4 x 1/3 = 4/3

Hence, the length of latus rectum is 4/3 units.
Discussion
57.

Find the length of the latus rectum of the parabola y2 = 4ax passing through the point (2, -6).

Answer»

Given equation of the parabola is y2 = 4ax and it passes through point (2, -6). 

Substituting x = 2 and y = -6 in y2 = 4ax, we get 

⇒ (-6)2 = 4a(2) 

⇒ 4a = 18 

∴ Length of latus rectum = 4a = 18 units
Discussion
58.

If e is the eccentricity of the ellipse x2/a2 + y2/b2 = 1(a < b), thenA. b2 = a2 (1 – e2)B. a2 = b2 (1 – e2)C. a2 = b2 (e2 – 1)D. b2 = a2 (e2 – 1)

Answer»

b2 = a2 (1 - e2)

Hence, Option A is correct.
Discussion