JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
Let the area enclosed by the $x$-axis, and the tangent and normal drawn to the curve $4x^3-3xy^2+6x^2-5xy-8y^2+9x+14=0$ at the point $(-2,3)$ be $A$. Then $8A$ is equal to ________.
Application of Integrals
Answer$170$
Solution
At $(-2,3)$ the equation gives $F=0$ (the point lies on the curve). $F_x=12x^2-3y^2+12x-5y+9=48-27-24-15+9=-9$, $F_y=-6xy-5x-16y=36+10-48=-2$. Slope $m=-F_x/F_y=-\frac92$. Tangent: $y-3=-\frac92(x+2)$ meets $y=0$ at $x=-\frac43$. Normal: slope $\frac29$, $y-3=\frac29(x+2)$ meets $y=0$ at $x=-\frac{31}{2}$. Triangle base $=\left|-\frac43+\frac{31}{2}\right|=\frac{85}{6}$, height $3$: $A=\frac12\cdot\frac{85}{6}\cdot3=\frac{85}{4}$. $8A=170$.