Question ID: #731
If the line $\alpha x+2y=1$, where $\alpha\in\mathbb{R}$, does not meet the hyperbola $x^{2}-9y^{2}=9$ then a possible value of $\alpha$ is:
- (1) 0.6
- (2) 0.8
- (3) 0.5
- (4) 0.7
Solution:
Equation of the line is $2y = 1 – \alpha x \Rightarrow y = \frac{1-\alpha x}{2}$.
Substitute $y$ into the equation of the hyperbola $x^2 – 9y^2 = 9$:
$$ x^2 – 9\left( \frac{1-\alpha x}{2} \right)^2 = 9 $$
$$ x^2 – \frac{9}{4}(1 + \alpha^2 x^2 – 2\alpha x) = 9 $$
Multiply by 4 to clear the denominator:
$$ 4x^2 – 9(1 + \alpha^2 x^2 – 2\alpha x) = 36 $$
$$ 4x^2 – 9 – 9\alpha^2 x^2 + 18\alpha x – 36 = 0 $$
$$ (4 – 9\alpha^2)x^2 + 18\alpha x – 45 = 0 $$
For the line not to meet the hyperbola, the quadratic equation must have no real roots.
Discriminant $D < 0$: $$ (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) < 0 $$ $$ 324\alpha^2 + 180(4 - 9\alpha^2) < 0 $$
Divide by 36:
$$ 9\alpha^2 + 5(4 – 9\alpha^2) < 0 $$ $$ 9\alpha^2 + 20 - 45\alpha^2 < 0 $$ $$ 20 - 36\alpha^2 < 0 $$ $$ 36\alpha^2 > 20 \Rightarrow \alpha^2 > \frac{20}{36} = \frac{5}{9} $$
$$ |\alpha| > \frac{\sqrt{5}}{3} $$
Approximating the value:
$$ \frac{\sqrt{5}}{3} \approx \frac{2.236}{3} \approx 0.745 $$
Checking the options:
(1) $0.6 < 0.745$ (2) $0.8 > 0.745$ (Correct)
(3) $0.5 < 0.745$ (4) $0.7 < 0.745$
Ans. (2)
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