Question

From the point $P(1,\sqrt{3})$ on the circle $x^2+y^2=$ a tangent is drawn to the hyperbola $\frac{x^2}{4}-\frac{y^2}{1}=1$ which meets its transverse axis at $Q$. From $Q$ a line is drawn parallel to conjugate axis, which cuts the hyperbola at $R$ above the x-axis, then $PR$ equals.

Answer 1

For the given hyperbola, we find that $a=2,b=1$

Let the equation of the tangent be $y=mx\pm \sqrt{a^2{m}^2+b^2}$ which is the standard form of the equation of the tangent to a hyperbola through a point.

It passes through the point $P=(1,\sqrt{3})$. So, we have:

$\sqrt{3}=m\pm \sqrt{4m^2+1}$

On solving, we get $m=1-\frac{1}{\sqrt{3}}=0.423$

The tangent cuts the x-axis at the point Q whose coordinates are given by:

$(-\frac{\sqrt{a^2{m}^2+b^2}}{m},0)=(-3.1,0)$

The point R lies above the x-axis and on the left branch of the hyperbola, so:

So, substituting $x=-\frac{\sqrt{a^2{m}^2+b^2}}{m}$ into the equation of the hyperbola, and solving for the positive value of y, we get:

$y=\frac{b^2}{am}=1.183$.

Thus, the coordinates of the point R are: $(-3.1,1.183)$

So, the length of the line PR is $\sqrt{(\sqrt{3}-1.183{)}^2+(1+3.1{)}^2}=4.14$