Question

Equation of tangent to the hyperbola $2x^2−3y^2=6$ which is parallel to the line $y=3x+4,$ is

• A. $y=3x−4$
• B. $y=3x−5$
• C. $y=3x+5$
• D. $y=3xÂ\pm 5$

Answer 1

Answer: (D)

Given hyperbola is
$2x^2−3y^2=6$
$⇒$ $\frac{x^2}{3}−\frac{y^2}{2}=1$
Here, $a^2=3,b^2=2$
Since, tangent is parallel to $y=3x+4$
$∴$ Here, $m=3$
Thus, tangent of hyperbola is
$y=mxÂ\pm \sqrt{a^2{m}^2−b^2}$
$⇒$ $y=3xÂ\pm \sqrt{3.9−2}$
$⇒$ $y=3xÂ\pm 5$