请帮我批改一下这篇作业吧~---**Question:** 7. 已知 F₁, F₂ 是椭圆 C: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a>b>0)\) 的两个焦点. P 为椭圆 C 上的一点. 且 \(PF_1 \perp PF_2\). 若 \(\triangle PF_1F_2\) 的面积为 9. 则 \(b = \underline{\hspace{1cm}}\). **Diagram Description:** * Type: Ellipse with horizontal and vertical coordinate axes. * Main Elements: * An ellipse centered at the origin, oriented horizontally. * Foci \(F_1\) and \(F_2\) are located on the horizontal axis inside the ellipse, symmetric about the origin. \(F_1\) is to the left of the origin, and \(F_2\) is to the right. * A point P is shown on the upper part of the ellipse. * Line segments \(PF_1\) and \(PF_2\) are drawn, forming the triangle \(PF_1F_2\). * A right angle symbol is indicated at point P, denoting that the angle between \(PF_1\) and \(PF_2\) is 90 degrees (\(\angle F_1PF_2 = 90^\circ\)). * An angle labeled \(\theta\) is shown at point P, within \(\triangle PF_1F_2\). * Horizontal and vertical axes pass through the center of the ellipse. **Handwritten Calculations and Notes:** * \(|PF_1| + |PF_2| = 2a\) * \((|PF_1| + |PF_2|)^2 = (2a)^2 = 4a^2\) * \(|PF_1|^2 + |PF_2|^2 + 2|PF_1||PF_2| = 4a^2\) * \(|F_1F_2|^2 + 2|PF_1||PF_2| = 4a^2\) (Applying Pythagorean theorem: \(|PF_1|^2 + |PF_2|^2 = |F_1F_2|^2\) since \(PF_1 \perp PF_2\)) * \((2c)^2 + 2|PF_1||PF_2| = 4a^2\) (Where \(|F_1F_2| = 2c\)) * \(4c^2 + 2|PF_1||PF_2| = 4a^2\) * \(2|PF_1||PF_2| = 4a^2 - 4c^2 = 4(a^2 - c^2)\) * Since \(a^2 - c^2 = b^2\) for an ellipse, \(2|PF_1||PF_2| = 4b^2\) * \(|PF_1||PF_2| = 2b^2\) * Area of \(\triangle PF_1F_2 = \frac{1}{2} |PF_1| |PF_2|\) (Since \(PF_1 \perp PF_2\)) * Area \( = \frac{1}{2} (2b^2) = b^2\) * Given Area = 9. Therefore, \(b^2 = 9\). * Formula noted: \(S_{\triangle PF_1F_2} = b^2 \cdot \tan(\frac{\theta}{2})\) * \( = b^2 \cdot \tan(45^\circ) = 9\) (Assuming \(\theta = 90^\circ\), so \(\theta/2 = 45^\circ\)) * \(= b^2 \cdot 1 = 9\) * Thus \(b^2 = 9\).