41. Let P be a point (other than origin) of intersection of the curves y² = 4ax and ay² = 4x³ such that the normals to the two curves meet x-axis at G₁ and G₂ resp. then, G₁ G₂ =
1.2a
2.4a
3.a
4.none
Correct ans 2.plz exp.
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Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{y}}^2=4\mathrm{ax}....\left(\mathrm{i}\right) \\ {\mathrm{ay}}^2=4{\mathrm{x}}^3...\left(\mathrm{ii}\right) \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{a}\left(4\mathrm{ax}\right)=4{\mathrm{x}}^3 \\ \Rightarrow {\mathrm{a}}^2\mathrm{x}={\mathrm{x}}^3 \\ \Rightarrow {\mathrm{x}}^3-{\mathrm{a}}^2\mathrm{x}=0 \\ \Rightarrow \mathrm{x}\left({\mathrm{x}}^2-{\mathrm{a}}^2\right)=0 \\ \mathrm{x}=0 \mathrm{is} \mathrm{not} \mathrm{to} \mathrm{be} \mathrm{considered}. \\ \Rightarrow {\mathrm{x}}^2-{\mathrm{a}}^2=0 \\ \Rightarrow {\mathrm{x}}^2={\mathrm{a}}^2 \\ \Rightarrow \mathrm{x}=\mathrm{a} \\ \mathrm{When} \mathrm{x}=\mathrm{a} \\ {\mathrm{y}}^2=4\mathrm{aa}=4{\mathrm{a}}^2 \\ \Rightarrow \mathrm{y}=\pm 2\mathrm{a} \\ \mathrm{Hence} \mathrm{we} \mathrm{have} \left(\mathrm{a},2\mathrm{a}\right) \mathrm{and} \left(\mathrm{a},-2\mathrm{a}\right) \mathrm{as} \mathrm{intersection} \mathrm{points}. \\ \mathrm{We} \mathrm{only} \mathrm{need} \mathrm{to} \mathrm{consider} \mathrm{one} \mathrm{point} \\ {\mathrm{y}}^2=4\mathrm{ax} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \Rightarrow 2\mathrm{y}.\frac{\mathrm{dy}}{\mathrm{dx}}=4\mathrm{a}...\left(\mathrm{iii}\right) \\ \mathrm{Put} \left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{a},2\mathrm{a}\right) \\ \Rightarrow 2\mathrm{a}.\frac{\mathrm{dy}}{\mathrm{dx}}=4\mathrm{a} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \\ \Rightarrow -\frac{\mathrm{dx}}{\mathrm{dy}}=-\frac{1}{2}=\mathrm{Slope} \mathrm{of} \mathrm{first} \mathrm{normal} \\ \mathrm{Equation} \mathrm{of} \mathrm{first} \mathrm{normal} \mathrm{passing} \mathrm{through} \left(\mathrm{a},2\mathrm{a}\right) \mathrm{is} \\ \mathrm{y}-2\mathrm{a}=-\frac{1}{2}\left(\mathrm{x}-\mathrm{a}\right) \\ \mathrm{For} \mathrm{x} \mathrm{axis}, \mathrm{y}=0 \\ -2\mathrm{a}=-\frac{1}{2}\left(\mathrm{x}-\mathrm{a}\right) \\ \Rightarrow \mathrm{x}-\mathrm{a}=4\mathrm{a} \\ \Rightarrow \mathrm{x}=5\mathrm{a} \\ \mathrm{Hence} {\mathrm{G}}_1\left(5\mathrm{a},0\right) \\ {\mathrm{ay}}^2=4{\mathrm{x}}^3 \\ \Rightarrow 2\mathrm{ay}.\frac{\mathrm{dy}}{\mathrm{dx}}=12{\mathrm{x}}^2 \\ \mathrm{On} \mathrm{putting} \left(\mathrm{a},2\mathrm{a}\right) \mathrm{we} \mathrm{get} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{4}=\mathrm{Slope} \mathrm{of} \mathrm{second} \mathrm{normal} \\ \mathrm{Equation} \mathrm{of} \mathrm{second} \mathrm{normal} \mathrm{passing} \mathrm{through} \left(\mathrm{a},2\mathrm{a}\right) \mathrm{is} \\ \mathrm{y}-2\mathrm{a}=-\frac{1}{4}\left(\mathrm{x}-\mathrm{a}\right) \\ \mathrm{For} \mathrm{x} \mathrm{axis}, \mathrm{y}=0 \\ -2\mathrm{a}=-\frac{1}{4}\left(\mathrm{x}-\mathrm{a}\right) \\ \Rightarrow \mathrm{x}-\mathrm{a}=8\mathrm{a} \\ \Rightarrow \mathrm{x}=9\mathrm{a} \\ \mathrm{Hence} {\mathrm{G}}_2\left(9\mathrm{a},0\right) \\ \Rightarrow {\mathrm{G}}_1{\mathrm{G}}_2=9\mathrm{a}-5\mathrm{a}=4\mathrm{a} \end{gathered}$$

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