Question no. 24 Part B
Q.24. (B) If line 2x - by + 1 = 0 intersects the curve $2x^2-b{y}^2+\left(2b-1\right)xy-x-by=0$ at points A & B and AB subtends a right angle at origin, then value of $b+b^2$ is equal to

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$$\begin{gathered} \mathrm{This} \mathrm{is} \mathrm{question} \mathrm{of} \mathrm{homogenization} \\ \mathrm{We} \mathrm{have} \\ 2\mathrm{x}-\mathrm{by}+1=0 \\ \Rightarrow 1=\mathrm{by}-2\mathrm{x}...\left(\mathrm{i}\right) \\ \mathrm{Now} \\ 2{\mathrm{x}}^2-{\mathrm{by}}^2+\left(2\mathrm{b}-1\right)\mathrm{xy}-\mathrm{x}-\mathrm{by}=0 \\ 2{\mathrm{x}}^2-{\mathrm{by}}^2+\left(2\mathrm{b}-1\right)\mathrm{xy}-\mathrm{x}\left(1\right)-\mathrm{by}\left(1\right)=0 \\ \mathrm{Use} \left(\mathrm{i}\right) \\ 2{\mathrm{x}}^2-{\mathrm{by}}^2+\left(2\mathrm{b}-1\right)\mathrm{xy}-\mathrm{x}\left(\mathrm{by}-2\mathrm{x}\right)-\mathrm{by}\left(\mathrm{by}-2\mathrm{x}\right)=0 \\ \Rightarrow 2{\mathrm{x}}^2-{\mathrm{by}}^2+\left(2\mathrm{b}-1\right)\mathrm{xy}-\mathrm{bxy}+2{\mathrm{x}}^2-{\mathrm{b}}^2{\mathrm{y}}^2+2\mathrm{bxy}=0 \\ \Rightarrow 4{\mathrm{x}}^2+{\mathrm{y}}^2\left(-\mathrm{b}-{\mathrm{b}}^2\right)+\left(2\mathrm{b}-1\right)\mathrm{xy}-\mathrm{bxy}+2\mathrm{bxy}=0 \\ \mathrm{As} \mathrm{line} \mathrm{subtend} \mathrm{right} \mathrm{angle} \mathrm{at} \mathrm{origin}, \mathrm{hence} \\ \mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^2+\mathrm{Coefficient} \mathrm{of} {\mathrm{y}}^2=0 \\ 4-\mathrm{b}-{\mathrm{b}}^2=0 \\ \Rightarrow \mathrm{b}+{\mathrm{b}}^2=4 \end{gathered}$$

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