JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q74

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MathematicsSingle correct+4 / -1Question 74 of 90

Let the area enclosed by the solution curve $y=y(x)$ of the differential equation $\dfrac{dy}{dx}+\dfrac{x-2}{y-b}=2$ , b > 0, y(5) = 0 is 13π. Let $y=y(x)$ intersect the y-axis at the points P and Q. If the tangents to $y=y(x)$ at P and Q meet at T, then the area of ΔPTQ is

13.5

14.5

Answer and worked solution

Official answer: A

The curve. A closed solution curve of area 13π comes from $\frac{dy}{dx}=-\frac{x-2}{y-b}$ (right side 0; the printed 2 does not give a closed curve): $(x-2)^2+(y-b)^2=13$ . Then $y(5)=0$ gives $9+b^2=13$ , $b=2$ .

Points. On $x=0$ : $(y-2)^2=9$ , so P(0, 5) and Q(0, −1). The tangent at P is $-2x+3y=15$ ; by symmetry T lies on $y=2$ , at $x=-4.5$ .

Area $\frac12\times6\times4.5=13.5$ .

(A)

Topics:Variable separable Tangents and normals

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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 74