Question

If P\(\left(\frac{\pi}{4}\right)\) and Q\(\left(\frac{3\pi}{4}\right)\) are two points on the hyperbola $4x^2-y^2-8x-2y-13=0$ in parametric form, then the distance between P and Q is:

• A. $4\sqrt{6}$
• B. $10$
• C. $8\sqrt{3}$
• D. $5$

Hint:

When working with conics in parametric form, it's often easier to first convert the given equation to standard form, identify the key parameters, and then use the appropriate parametric equations.

Answer 1

The Correct Option is A

The hyperbola is given by $4x^2-y^2-8x-2y-13=0$ First, let's rewrite this equation in standard form by completing the squares: For the $x$ terms: $4x^2-8x = 4(x^2-2x) = 4(x^2-2x+1-1) = 4(x-1)^2-4$ For the $y$ terms: $-y^2-2y = -(y^2+2y) = -(y^2+2y+1-1) = -(y+1)^2+1$ Therefore, $4x^2-y^2-8x-2y-13 = 4(x-1)^2-(y+1)^2-4+1-13 = 4(x-1)^2-(y+1)^2-16 = 0$ This gives us: $4(x-1)^2-(y+1)^2 = 16$ Dividing by 16: $\frac{(x-1)^2}{4}-\frac{(y+1)^2}{16} = 1$ This is a hyperbola with center at $(1,-1)$, $a=2$ and $b=4$. The parametric equations for this hyperbola are: $x = 1 + 2\sec t$ $y = -1 + 4\tan t$ For point P at $t = \frac{\pi}{4}$: $x_P = 1 + 2\sec\frac{\pi}{4} = 1 + 2\sqrt{2} = 1 + 2.83... \approx 3.83$ $y_P = -1 + 4\tan\frac{\pi}{4} = -1 + 4 \cdot 1 = 3$ For point Q at $t = \frac{3\pi}{4}$: $x_Q = 1 + 2\sec\frac{3\pi}{4} = 1 + 2(-\sqrt{2}) = 1 - 2.83... \approx -1.83$ $y_Q = -1 + 4\tan\frac{3\pi}{4} = -1 + 4 \cdot (-1) = -5$ Now we can find the distance between P and Q: $d = \sqrt{(x_Q-x_P)^2 + (y_Q-y_P)^2}$ $d = \sqrt{(-1.83-3.83)^2 + (-5-3)^2}$ $d = \sqrt{(-5.66)^2 + (-8)^2}$ $d = \sqrt{32 + 64}$ $d = \sqrt{96}$ $d = 4\sqrt{6}$ Therefore, the distance between points P and Q is $4\sqrt{6}$.