Question

If the distance between two parallel tangents drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{49}=1$ is $2$ unit, then which of the following is(are) correct?

• A. Quadrilateral formed by the points of contact of the tangent is rectangle.
• B. Slope of one pair of parallel tangent is $\frac{5}{2}$
• C. Slope of one pair of parallel tangent is $-\frac{5}{2}$
• D. area formed by the points of contact of the tangent is $180$ sq. units

Answer 1

Answer: (A), (B), (C)

Slope of one pair of parallel tangent is $-\frac{5}{2}$

For the given hyperbola the equation of parallel tangents is
$y=mx\pm \sqrt{9m^2-49}$
Distance between parallel tangents
$2=\frac{|c_1-c_2|}{\sqrt{1+m^2}}$
$\Rightarrow \frac{2\sqrt{9m^2-49}}{\sqrt{1+m^2}}=2$
$\Rightarrow 9m^2-49=1+m^2$
$\Rightarrow 8m^2=50$
$\Rightarrow m=\pm \frac{5}{2}$
Hence $c_1=\frac{\sqrt{29}}{2},c_2=-\frac{\sqrt{29}}{2}$
$\therefore$ point of contacts of the tangents will be
$(\frac{45}{\sqrt{29}},\frac{98}{\sqrt{29}}),(-\frac{45}{\sqrt{29}},-\frac{98}{\sqrt{29}}),(\frac{45}{\sqrt{29}},-\frac{98}{\sqrt{29}}),(-\frac{45}{\sqrt{29}},\frac{98}{\sqrt{29}})$
Hence quadrilateral formed by the point of contact is a rectangle with sides $\frac{90}{\sqrt{29}}$ unit and $\frac{196}{\sqrt{29}}$ unit
So, required area
$=\frac{90}{\sqrt{29}}\times \frac{196}{\sqrt{29}}=\frac{17640}{29}$ sq. units