Question

From point $(2,2)$ tangents are drawn to the hyperbola $\frac{x2}{16}-\frac{y2}{9}=1$ then the point of contacts lie in

• A. I and II are quadrants
• B. I and IV quadrants
• C. I and III quadrants
• D. III and IV quadrants

Answer 1

The correct option is D III and IV quadrants

Let $m_1\&m_2$ be slopes of tangents to $\frac{x^2}{16}-\frac{y^2}{9}=1$ from $(2,2)$

$\therefore m_1\&m_2$ are roots of the equation -

$\Rightarrow (x_1-a^2)m^2-2x_1{y}_{1}m+(y_1^2+b^2)=0$

$\Rightarrow (2-16)m^2-8m+(4+9)=0$

$\Rightarrow -14m^2-8m+13=0$

$\Rightarrow m_1=0.7194$

$\Rightarrow m_2=-1.2908$

$\therefore$ Equation of tangents are -

$\Rightarrow (y-2)=0.7194(x-2);(y-2)=-1.2908(x-2)$

$y=0.7194x+0.5162;y=-1.2908x+4.5816$

Putting these values in equation of hyperbola one by one. We find that points of contact lie on $3rd$ and $4th$ quadrants.

Hence, the answer is III and IV quadrants.