Question

Find the midpoint of the line joining the points $P$ on the rectangular hyperbola $x^2-y^2=16$ with eccentric angle $\frac{\pi }{4}$ and the point $Q$ on the conjugate rectangular hyperbola $x^2-y^2=-16$ with eccentric angle $\frac{\pi }{3}$.

• A. $(2\sqrt{2}+4,2-2\sqrt{3})$
• B. $(2\sqrt{2}-4,2+2\sqrt{3})$
• C. $(2\sqrt{2}+3,2+2\sqrt{3})$
• D. $(2(\sqrt{2}+\sqrt{3}),6)$

Answer 1

The correct option is D $(2(\sqrt{2}+\sqrt{3}),6)$

Given: $P$ on the rectangular hyperbola $x^2-y^2=16$;
$Q$ on the conjugate rectangular hyperbola $x^2-y^2=-16$

Eccentric angle of $P$: $\frac{\pi }{4}$

Eccentric angle of $Q$: $\frac{\pi }{3}$

To find: Midpoint of the line joining the points $P$ and $Q$

Step-$1$: Find the coordinates of $P$ and $Q$

Step-$2$: Using the midpoint formula, find the midpoint of the line joining $P$ and $Q$

We know that, the parametric coordinates of a point on a standard rectangular hyperbola $x^2-y^2=a^2$ is $(a\sec \theta ,a\tan \theta )$.

On comparing the given hyperbola equation with the standard rectangular hyperbola equation, we have $a^2=16$

$\Rightarrow a=4$

Substituting $a=4$ and $\theta =\frac{\pi }{4}$ in the coordinates $(a\sec \theta ,a\tan \theta )$.

$\Rightarrow (4\sec (\frac{\pi }{4}),4\tan (\frac{\pi }{4}))$

$\Rightarrow (4\times \sqrt{2},4\times 1)$

$\Rightarrow P(4\sqrt{2},4)$

We know that, the parametric coordinates of a point on a standard conjugate rectangular hyperbola $x^2-y^2=-a^2$ is $(a\tan \theta ,a\sec \theta )$.

On comparing the given conjugate rectangular hyperbola equation with the standard conjugate rectangular hyperbola equation, we have $a^2=16$

$\Rightarrow a=4$

Substituting $a=4$ and $\theta =\frac{\pi }{3}$ in the coordinates $(a\tan \theta ,a\sec \theta )$.

$\Rightarrow (4\tan (\frac{\pi }{3}),4\sec (\frac{\pi }{3}))$

$\Rightarrow (4\times \sqrt{3},4\times 2)$

$\Rightarrow Q(4\sqrt{3},8)$

The line joining the points $P$ and $Q$ will be $PQ$.

Midpoints of the line $PQ$ will be $(\frac{P_x+Q_x}{2},\frac{P_y+Q_y}{2})$.

We have,
$P_x=4\sqrt{2}{Q}_x=4\sqrt{3}{P}_y=4Q_y=8$

$\Rightarrow (\frac{4\sqrt{2}+4\sqrt{3}}{2},\frac{4+8}{2})$

$\Rightarrow (2(\sqrt{2}+\sqrt{3}),6)$