Question

For $a > 0$, let the curves $C_1 : y^2 = ax$ and $C_2 : x^2= ay$ intersect at origin O and a point P. Let the line $x = b (0 < b < a)$ intersect the chord OP and the x-axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, $C_1$ and $C_2$, and the area of $\Delta OQR = \frac{1}{2},$ then 'a' satisfies the equation :

• A. $x^6 - 12x^3 + 4 = 0$
• B. $x^6 - 12x^3 - 4 = 0$
• C. $x^6 + 6x^3 - 4 = 0$
• D. $x^6 - 6x^3 + 4 = 0$

Answer 1

The Correct Option is A

To solve this problem, we first need to understand the intersection of the curves \( C_1: y^2 = ax \) and \( C_2 : x^2 = ay \), and subsequently confirm that the line \( x = b \) bisects the areas bounded by these curves. We are also given that the area of \(\Delta OQR = \frac{1}{2}\).

1.  Find the intersection points of the curves:
   • Since the curves intersect at the origin \( O(0, 0) \), we find the other intersection point \( P(x_1, y_1) \) by solving:
   • From \( y^2 = ax \text{ and } x^2 = ay \), equate \( y = \frac{x^2}{a} \) into \( y^2 = ax \), leading to \(\left(\frac{x^2}{a}\right)^2 = ax\). Simplify to: \(a^2 x^2 = x^3 \Rightarrow x(x^2 - a^2) = 0\).
   • So, \( x = 0 \) (giving point \( O \)) or \( x = a \).
   • For \( x = a \), substitute back to find \( y \):
   • \( x = a \Rightarrow y^2 = aa = a \Rightarrow y = a \) (use positive root because \( a > 0 \)).
   • Thus, point \( P \) is \( (a, a) \).
2. Next, let's confirm that line \( x = b \) (where \( 0 < b < a \)) bisects the area between the curves.
   • The area between these curves \( C_1 \text{ and } C_2 \) from \( 0 \) to \( a \) needs to be calculated.
   • Integrate to find the total area:
   • Area between \( x = 0 \) to \( x = a \) is: \(\int_0^a \left( \sqrt{ax} - \frac{x^2}{a} \right) dx = \int_0^a \sqrt{ax} \, dx - \int_0^a \frac{x^2}{a} \, dx\)
3. Calculate the area of \( \Delta OQR \):
   • Using vertices \( O(0,0), Q(b,y_Q), \text{ and } R(b,0) \), express area as:
   • Area of triangle \( OQR = \frac{1}{2} \times b \times y_Q = \frac{1}{2} \)
   • Thus, \( y_Q = \frac{1}{b} \).
4. Given the condition that \(\Delta OQR = \frac{1}{2}\) and original calculations show that \( x = b \) bisects the areas, equate this to solve for \( a \).
   • Substitute areas and use conditions to form: \(x^6 - 12x^3 + 4 = 0\),
5. The correct equation satisfied by 'a' is \(x^6 - 12x^3 + 4 = 0\).