Test Index

Hyperbola

Section: Mathematics

© examsnet.com

Question : 12 of 98

Marks: +1, -0

If the line

\alpha x+2 y=1

\alpha \in R

, does not meet the hyperbola

x^2-9 y^2=9

, then a possible value of

\alpha

[22 Jan 2026 Shift 1]

0.6
0.7
0.8
0.5

Solution:

Concept:

If a line does not meet a hyperbola, the quadratic obtained after substitution has a negative discriminant.

Explanation:

Line:

\alpha x + 2y = 1

⇒

y = \frac{1 - \alpha x}{2}

.

Hyperbola:

x^2 - 9y^2 = 9

.

Substitute

y

:

x^2 - 9\left(\frac{1 - \alpha x}{2}\right)^2 = 9

.

Multiply by

4

:

4x^2 - 9(1 - 2\alpha x + \alpha^2 x^2) = 36

.

Simplify:

(4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0

.

For no intersection, discriminant

\Delta < 0

:

\Delta = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) = 324\alpha^2 + 720 - 1620\alpha^2 = 720 - 1296\alpha^2

.

Set

\Delta < 0

:

720 - 1296\alpha^2 < 0

⇒

\alpha^2 > \frac{720}{1296} = \frac{5}{9} \approx 0.5556

.

Thus

\alpha > \sqrt{\frac{5}{9}} \approx 0.745

or

\alpha < -0.745

.

From the options, only

\alpha = 0.8

satisfies

\alpha^2 > 0.5556

.

Answer:

C. 0.8

© examsnet.com

Go to Question:

Prev Question

Next Question