Let $L_1$ be the length of the common chord of the curves $x^2+y^2 = 9$ and $y^2 = 8x,$ and $L_2$ be the length of the latus rectum of $y^2 = 8x,$ then :

Question

The area bounded by $ y - 1 = |x| $ and $ y + 1 = |x| $ is: (a) $ \frac{1}{2} $

Answer 1

To solve this problem, we need to determine the lengths $L_1$ and $L_2$, and then compare them.

The first curve given is a circle: $x^2 + y^2 = 9$. This is a circle centered at the origin with radius $3$.
The second curve given is a parabola: $y^2 = 8x$. This is a standard parabola that opens to the right.
To find the length of the common chord $L_1$, we find points of intersection of these curves by solving both equations simultaneously:
- Substitute $y^2 = 8x$ into the circle equation:
  $x^2 + 8x = 9$
- Rearrange into a standard quadratic equation:
  $x^2 + 8x - 9 = 0$
- Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 8$, $c = -9$:
- Solve for $x$:
  $x = \frac{-8 \pm \sqrt{64 + 36}}{2}$
  $x = \frac{-8 \pm 10}{2}$
  $x = 1$ or $x = -9$ (only $x = 1$ is valid as $y^2 \ge 0$)
- For $x = 1$, $y^2 = 8 \times 1 = 8, \quad y = \pm 2\sqrt{2}$
- The points of intersection are therefore $(1, 2\sqrt{2})$ and $(1, -2\sqrt{2})$.
The length of the common chord $L_1$ can be found via the distance formula between these points:
$L_1 = \sqrt{(1 - 1)^2 + (2\sqrt{2} - (-2\sqrt{2}))^2} = \sqrt{(4\sqrt{2})^2} = 4\sqrt{2}$
The length of the latus rectum $L_2$ for the parabola $y^2 = 4ax$ is given by $4a$. Here, $4a = 8$ implies $a = 2$, so $L_2 = 8$.
Now, comparing $L_1 = 4\sqrt{2}$ and $L_2 = 8$:
Since $4\sqrt{2} \approx 5.656$, we find that $L_1 < L_2$.

Hence, the correct answer is $L_1 < L_2$.

The Correct Option is C