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If a line $ax+y=1$ does not intersect the hyperbola $x^{2}-9y^{2}=9$, then possible value of $a$ is :
APPLY COMPETENCY 4 Marks
Concept Application
50%
Calculation / Logic
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Target Level
MEDIUM
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APPLY COMPETENCY MEDIUM

Q: If a line $ax+y=1$ does not intersect the hyperbola $x^{2}-9y^{2}=9$, then possible value of $a$ is :

Question Analysis & Solution

Step-by-Step Solution

The equation of the hyperbola is $x^2 - 9y^2 = 9$. We can rewrite this as $\frac{x^2}{9} - y^2 = 1$.

The equation of the line is $ax + y = 1$, which can be rewritten as $y = 1 - ax$.

Substitute the equation of the line into the equation of the hyperbola:

$x^2 - 9(1 - ax)^2 = 9$

$x^2 - 9(1 - 2ax + a^2x^2) = 9$

$x^2 - 9 + 18ax - 9a^2x^2 = 9$

$(1 - 9a^2)x^2 + 18ax - 18 = 0$

For the line not to intersect the hyperbola, the quadratic equation must have no real solutions. This means the discriminant must be negative.

The discriminant is $D = b^2 - 4ac = (18a)^2 - 4(1 - 9a^2)(-18)$

$D = 324a^2 + 72(1 - 9a^2) = 324a^2 + 72 - 648a^2 = 72 - 324a^2$

For no intersection, $D < 0$

$72 - 324a^2 < 0$

$324a^2 > 72$

$a^2 > \frac{72}{324} = \frac{2}{9}$

$|a| > \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3} \approx \frac{1.414}{3} \approx 0.471$

So, $a > 0.471$ or $a < -0.471$

From the options, the possible value of $a$ is 0.5.

Correct Answer: 0.5

AI Suggestion: Option D
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