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1

### WB JEE 2009

MCQ (Single Correct Answer)

The angle between the line joining the foci of an ellipse to one particular extremity of the minor axis is 90$$^\circ$$. The eccentricity of the ellipse is

A
$${1 \over 8}$$
B
$${1 \over {\sqrt 3 }}$$
C
$$\sqrt {{2 \over 3}}$$
D
$$\sqrt {{1 \over 2}}$$

## Explanation

Let the equation of the ellipse is $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

coordinates of foci are (ae, 0), ($$-$$ae, 0)

coordinates of extremity of minor axis are (0, b)

Slopes of lines joining foci to extremity are $$- {b \over {ae}},{b \over {ae}}$$

but lines are at right angle

$$\therefore$$ $$- {b \over {ae}}.{b \over {ae}} = - 1 \Rightarrow {b^2} = {a^2}{e^2}$$ .... (i)

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

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

$$\Rightarrow e = {1 \over {\sqrt 2 }}$$

2

### WB JEE 2009

MCQ (Single Correct Answer)

The line y = 2t2 intersects the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ in real point if

A
$$|t| \le 1$$
B
$$|t| < 1$$
C
$$|t| > 1$$
D
$$|t| \ge 1$$

## Explanation

Keep y = 2t2 in $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$, we get

$${{{x^2}} \over 9} + {{{{(2{t^2})}^2}} \over 4} = 1 \Rightarrow {x^2} = 9(1 - {t^4})$$

If line intersects the ellipse in real points, then x has real roots

So $${x^2} \ge 0 \Rightarrow 9(1 - {t^4}) \ge 0 \Rightarrow (1 - {t^2})(1 + {t^2}) \ge 0$$

$$\Rightarrow 1 - {t^2} \ge 0 \Rightarrow |t| \ge 1$$ [$$\because$$ $$1 + {t^2} > 0$$]

3

### WB JEE 2009

MCQ (Single Correct Answer)

The total number of tangents through the point (3, 5) that can be drawn to the ellipse 3x2 + 5y2 = 32 and 25x2 + 9y2 = 450 is

A
0
B
2
C
3
D
4

## Explanation

$$\because$$ 3 . 32 + 5 . 52 $$-$$ 32 > 0

$$\therefore$$ From (3, 5) we can draw 2 tangents to the ellipse

3x2 + 5y2 = 32. Again 25 . 32 + 9 . 52 $$-$$ 450 = 0

$$\therefore$$ From (3, 5) we can draw only 1 tangent to the ellipse

25x2 + 9y2 = 450 $$\therefore$$ Required number of tangents = 2 + 1 = 3

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