Question

Find the angle between the tangents draws from the point (5,3) to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Given point is P(5,3)

Hyperbpla $s=\frac{x^2}{25}-\frac{y^2}{9}=1=0$

$\because S_1=\frac{x^2}{25}-\frac{y^2}{9}-1=-1<0$

$\Rightarrow$ Point P$\equiv$ (5,3) lies outside the hyperbola

$\therefore$ Two tangents can be drawn from the point p(5,3) and

equation of pair of pair of tangents is $S{S}_1={\tau }^2$

$(\frac{x^2}{25}-\frac{y^2}{9}-1)(-1)={(\frac{5x}{25}-\frac{2y}{9}-1)}^2$

$\frac{-x^2}{25}+\frac{y^2}{9}+1=\frac{x^2}{25}+\frac{y^2}{9}+1-\frac{-2xy}{15}+\frac{2y}{3}-\frac{2x}{5}$

$\frac{2x^2}{25}-\frac{2xy}{15}+\frac{2y}{3}-\frac{2x}{5}=0$

$3x^2-5xy+25y-15x=0$

Equation of pair of tangents

$3x^2-5xy-15x+25y=0$

Comparing the given equation with standard pair of straight line equation

$a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0$

Tangent angle between these line $tan\theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$=\frac{2\sqrt{{\left(\frac{5}{2}\right)}^2-0}}{3+0}$

$=\frac{2\times \frac{5}{2}}{3}$

$\theta =ta{n}^{-1}\left(\frac{5}{3}\right)$