© All Rights reserved @ LearnWithDash
Step 1: Find the coordinates of point P
The point P lies in the first quadrant at the intersection of the circle
x^2 + y^2 = 3 and the parabola x^2 = 2y .
From x^2 = 2y, we get y = \frac{x^2}{2}.
Substitute y = \frac{x^2}{2} into the circle equation
x^2 + y^2 = 3. That becomes:
x^2 + \left(\frac{x^2}{2}\right)^2 = 3
\quad\Longrightarrow\quad x^2 + \frac{x^4}{4} = 3.
Multiply both sides by 4 to clear the fraction:
4x^2 + x^4 = 12
\quad\Longrightarrow\quad x^4 + 4x^2 - 12 = 0.
Let t = x^2. The equation becomes
t^2 + 4t - 12 = 0.
Solve this quadratic in t. The discriminant is
\Delta = 4^2 + 4 \cdot 12 = 16 + 48 = 64.
Hence,
t = \frac{-4 \pm \sqrt{64}}{2}
= \frac{-4 \pm 8}{2}.
Thus t = 2 or t = -6. Since x^2 \geq 0, we take t = 2.
This gives x^2 = 2 \implies x = \sqrt{2} (positive in the first quadrant).
Then y = \frac{x^2}{2} = \frac{2}{2} = 1.
Therefore, P is at \bigl(\sqrt{2},\,1\bigr).
Step 2: Determine the equation of the line L
We have L: \sqrt{2}\,x + y = \alpha, and it passes through P. Substitute
x = \sqrt{2} and y = 1 :
\sqrt{2}\cdot \sqrt{2} + 1 = \alpha
\quad\Longrightarrow\quad 2 + 1 = \alpha
\quad\Longrightarrow\quad \alpha = 3.
Hence, the line L is \sqrt{2}\,x + y = 3.
Step 3: Identify the centers Q₁ and Q₂ of the circles
Let the circles C_1 and C_2 both have radius 2\sqrt{3} and centers on the y-axis at
Q_1(0,\,y_1) and Q_2(0,\,y_2). The line L must be tangent to each circle,
so the perpendicular distance from each center to the line must be exactly 2\sqrt{3}.
Rewrite the line L as \sqrt{2}\,x + y - 3 = 0.
The distance from (0, y_0) to \sqrt{2}\,x + y - 3 = 0 is
\frac{|\sqrt{2}\cdot 0 + y_0 - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}}
= \frac{|y_0 - 3|}{\sqrt{2 + 1}}
= \frac{|y_0 - 3|}{\sqrt{3}}.
We require this to be 2\sqrt{3}, so
\frac{|y_0 - 3|}{\sqrt{3}} = 2\sqrt{3}
\quad\Longrightarrow\quad |y_0 - 3| = 6.
Thus y_0 - 3 = \pm 6 \implies y_0 = 9 \text{ or } y_0 = -3.
Hence Q_1 = (0,\,9) and Q_2 = (0,\,-3).
Step 4: Find the area of triangle P Q₁ Q₂ and then its square
The triangle has vertices
P\bigl(\sqrt{2},\,1\bigr),\, Q_1(0,\,9),\, Q_2(0,\,-3).
The area of a triangle with coordinates (x_1, y_1), (x_2, y_2), (x_3, y_3) is given by:
\text{Area}
= \frac{1}{2}
\Bigl|
x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)
\Bigr|.
Labeling P as (x_1, y_1),\,Q_1 as (x_2, y_2),\,Q_2 as (x_3, y_3):
x_1 = \sqrt{2},\,y_1 = 1,
x_2 = 0,\,y_2 = 9,
x_3 = 0,\,y_3 = -3.
Then y_2 - y_3 = 9 - (-3) = 12. Hence,
\text{Area}
= \frac{1}{2}\Bigl|\sqrt{2}\cdot 12 + 0 + 0\Bigr|
= \frac{1}{2} \times 12\sqrt{2}
= 6\sqrt{2}.
Therefore, the area of triangle PQ_1Q_2 is 6\sqrt{2}. Its square is
(6\sqrt{2})^2 = 72.
Final Answer
The square of the area of triangle PQ_1Q_2 is 72.
