JEE Main 2022 — Mathematics Calculus

Given that ${A}_{1}=2{A}_{2}$A1​=2A2​

Now, from the graph

$${A}_{1}+{A}_{2}=x\times y-2\times 4=xy-8$$A1​+A2​=x×y−2×4=xy−8

$$\Rightarrow \frac{3}{2}{A}_{1}=xy-8$$⇒23​A1​=xy−8

$$\Rightarrow {A}_{1}=\frac{2}{3}xy-\frac{16}{3}$$⇒A1​=32​xy−316​

$$\Rightarrow {\int }_{4}^{x}f(x)dx=\frac{2}{3}xy-\frac{16}{3}$$⇒∫4x​f(x)dx=32​xy−316​

$$\Rightarrow f(x)=\frac{2}{3}(x\frac{dy}{dx}+y)$$⇒f(x)=32​(xdxdy​+y)

$$\Rightarrow \frac{2}{3}x\frac{dy}{dx}=\frac{y}{3}$$⇒32​xdxdy​=3y​

$$\Rightarrow 2\int \frac{dy}{y}=\int \frac{dx}{x}$$⇒2∫ydy​=∫xdx​

$$\Rightarrow 2\mathrm{ln}y=\mathrm{ln}x+\mathrm{ln}c$$⇒2lny=lnx+lnc

$$\Rightarrow {y}^{2}=cx$$⇒y2=cx

As $f(4)=2\Rightarrow c=1$f(4)=2⇒c=1

So ${y}^{2}=x$y2=x is the given curve.

Given that the slope of normal $=-6$=−6

So the equation of normal will be

$$y=(-6x)-\frac{1}{2}(-6)-\frac{1}{4}{(-6)}^{3}$$y=(−6x)−21​(−6)−41​(−6)3

$$\Rightarrow y=-6x+3+54$$⇒y=−6x+3+54

$$\Rightarrow y+6x=57$$⇒y+6x=57

Only $(10,-4)$(10,−4) does not satisfy from the given options.