Question

Find the equation of tangent to the curve $4x^2-3y^2=24$ at the point whose ordinate is $2$ in the first quadrant.

• A. $\frac{x}{3}-\frac{y}{2}=1$
• B. None of the above.
• C. $\frac{x}{5}-\frac{y}{4}=1$
  
• D. $\frac{x}{2}-\frac{y}{4}=1$

Answer 1

The correct option is D $\frac{x}{2}-\frac{y}{4}=1$

The equation of hyperbola is $4x^2-3y^2=24$
$\Rightarrow \frac{4x^2}{24}-\frac{3y^2}{24}=1$
$\Rightarrow \frac{x^2}{6}-\frac{y^2}{8}=1$

Now let the absissa of the point of contact be $x_1$, so the point of contact becomes $(x_1,2)$.

Substituting $x=x_1$ and $y=2$ in the equation of the hyperbola, we get:
$\Rightarrow \frac{x_1^2}{6}-\frac{4}{8}=1$
$\Rightarrow x_1^2=9$
$\Rightarrow x_1=3,-3$

But $x_1\ne -3\because$ (Point is in the first quadrant)
$\therefore$ Point of contact $=(3,2)$.

Equation of tangent to a hyperbola at point P$(x_1,y_1)$ :
$\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$
So here, equation of tangent is $\frac{x.3}{6}-\frac{y.2}{8}-1=0$
$\frac{x}{2}-\frac{y}{4}=1$