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1

### WB JEE 2009

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

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

2

### WB JEE 2009

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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