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1

### WB JEE 2009

MCQ (Single Correct Answer)

The coordinates of the focus of the parabola described parametrically by x = 5t2 + 2, y = 10t + 4 are

A
(7, 4)
B
(3, 4)
C
(3, $$-$$4)
D
($$-$$7, 4)

## Explanation

Equation of parabola in parametric form is

x = 5t2 + 2, y = 10t + 4.

Eliminating t we get (y $$-$$ 4)2 = 20(x $$-$$ 2)

Comparing with Y2 = 4AX

y $$-$$ 4 = Y ..... (i)

x $$-$$ 2 = X ..... (ii)

4A = 20 $$\Rightarrow$$ A = 5

Focus of Y2 = 4AX is (A, 0) = (5, 0)

$$\therefore$$ y $$-$$ 4 = 0, x $$-$$ 2 = 5 or y = 4, x = 7

$$\therefore$$ Coordinates of focus of parabola are (7, 4).

2

### WB JEE 2008

MCQ (Single Correct Answer)

The latus rectum of an ellipse is equal to one-half of its minor axis. The eccentricity of the ellipse is

A
$${1 \over {\sqrt 6 }}$$
B
$${{\sqrt 3 } \over 2}$$
C
$${{\sqrt 3 } \over 4}$$
D
$${1 \over 2}$$

## Explanation

Latus rectum of ellipse = $${1 \over 2}$$ (minor axis) (given)

$$\Rightarrow {{2{b^2}} \over a} = {1 \over 2}(2b)$$

$$\Rightarrow 2{b^2} = ab \Rightarrow 2b = a$$ ..... (i)

Now, $${b^2} = {a^2}(1 - {e^2})$$

$${{{a^2}} \over 4} = {a^2}(1 - {e^2})$$

$$1 - {e^2} = {1 \over 4} \Rightarrow e = {{\sqrt 3 } \over 2}$$ (as $$-$$ve value of e is not possible)

3

### WB JEE 2008

MCQ (Single Correct Answer)

The equation of the ellipse having vertices at ($$\pm$$ 5, 0) and foci ($$\pm$$ 4, 0) is

A
$${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$
B
$$9{x^2} + 25{y^2} = 225$$
C
$${{{x^2}} \over 9} + {{{y^2}} \over {25}} = 1$$
D
$$4{x^2} + 5{y^2} = 20$$

## Explanation

Vertices $$\to$$ ($$\pm$$ a, 0) = ($$\pm$$ 5, 0) (given)

$$\therefore$$ a = $$\pm$$ 5 ..... (i)

Foci $$\to$$ ($$\pm$$ ae, 0) = ($$\pm$$ 4, 0) (given)

$$\therefore$$ ae = 4 ...... (ii)

From (i) and (ii), e = 4/5

Now, $${b^2} = {a^2}(1 - {e^2}) = {5^2}\left( {1 - {{\left( {{4 \over 5}} \right)}^2}} \right) = 25 \times {9 \over {25}} = 9$$

$$\therefore$$ b = $$\pm$$ 3

So, equation of ellipse will be

$${{{x^2}} \over {{5^2}}} + {{{y^2}} \over {{3^2}}} = 1 \Rightarrow {{{x^2}} \over {25}} + {{{y^2}} \over 9} = 1 \Rightarrow 9{x^2} + 25{y^2} = 225$$.

4

### WB JEE 2008

MCQ (Single Correct Answer)

If t is a parameter, then $$x = a\left( {t + {1 \over t}} \right)$$, $$y = b\left( {t - {1 \over t}} \right)$$ represents

A
An ellipse
B
A circle
C
A pair of straight lines
D
A hyperbola

## Explanation

$$x = a\left( {t + {1 \over t}} \right)$$, $$y = b\left( {t - {1 \over t}} \right)$$

or, $$t + {1 \over t} = {x \over a}$$ ..... (1)

$$t - {1 \over t} = {y \over b}$$ ....... (2)

Since $${\left( {t + {1 \over t}} \right)^2} - {\left( {t - {1 \over t}} \right)^2} = 4$$ (using 1 and 2)

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 4 \Rightarrow {{{x^2}} \over {4{a^2}}} - {{{y^2}} \over {4{b^2}}} = 1$$

This is equation of hyperbola.

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