Question

The points of contact of the line $y = x - 1$ with $3x^{2} - 4y^{2} = 12$ is

• A. (4, 3)
• B. (3, 4)
• C. (4, – 3)
• D. None of these

Answer 1

Answer: (A)

(4, 3)

Answer 2

The equation of line and hyperbola are $y = x - 1$ .....(i) and $3x^{2} - 4y^{2} = 12$ .....(ii)

From (i) and (ii), we get $3x^{2} - 4(x - 1)^{2} = 12$

⇒ $3x^{2} - 4(x^{2} - 2x + 1) = 12$ or $x^{2} - 8x + 16 = 0$ ⇒ $x = 4$

From (i), $y = 3$ so points of contact is (4, 3)

Trick : Points of contact are $\left( \pm \frac{a^{2}m}{\sqrt{a^{2}m^{2} - b^{2}}}, \pm \frac{b^{2}}{\sqrt{a^{2}m^{2} - b^{2}}} \right)$.

Here $a^{2} = 4$, $b^{2} = 3$ and $m = 1$. So the required points of contact is (4, 3)