$$ y^2 = x $$

दिए गए मध्यबिंदु वाले जीवा (Chord with given midpoint)

$$ \begin{aligned} & T=S_1 \\ & yk - \frac{x+h}{2} = k^2 - h \\ & 2ky - x - h = 2k^2 - 2h \end{aligned} $$

अब,

$$ \begin{aligned} & \left. A = \int_{y_1}^{y_2} \left(2ky - 2k^2 + h\right) - y^2\right) dy = \frac{4}{3} \\ & \left( ky^2 + (h - 2k^2) y - \frac{y^3}{3} \right)_{y_1}^{y_2} = \frac{4}{3} \\ & k(y_2^2 - y_1^2) + (h - 2k^2)(y_2 - y_1) - \frac{1}{3}(y_2^3 - y_1^3) = \frac{4}{3} \\ & (y_2 - y_1) [k - 2k + h - 2k^2 - \frac{1}{3}(4k^2 - 2k^2 + h)] = \frac{4}{3} \\ & 2(h - k^2)^{1/2} [2h - 2k^2] = 4 \\ & (h - k^2)^{3/2} = 1 \\ & x - y^2 = 1 \Rightarrow y^2 = x - 1 \end{aligned} $$

$$ \begin{aligned} & A = \int_1^4 (\sqrt{x} - \sqrt{x-1}) dx \\ & = \frac{2}{3} \left[ x^{3/2} - (x-1)^{3/2} \right]_1^4 = \frac{2}{3}[8 - 3\sqrt{3} - 1] \\ & = \frac{14}{3} - 2\sqrt{3} \end{aligned} $$

जीवा और परवलय का हल (Solving chord and parabola)

$$ \begin{aligned} & yk - \frac{y+h}{2} = k^2 - h \\ & 2ky - y^2 - h = 2k^2 - 2h \\ & y^2 - 2ky + 2k^2 - h = 0 \\ & y_1 + y_2 = 2k \\ & y_1 y_2 = 2k^2 - h \\ & y_2 - y_1 = \sqrt{4k^2 - 8k^2 + 4x} \\ & = 2\sqrt{h - k^2} \end{aligned} $$