JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q71

MathematicsSingle correct+4 / -1Question 71 of 90

The tangent at the point (2, −2) to the curve, $x^2y^2-2x=4(1-y)$ does not pass through the point :

\left(4,\dfrac13\right)

(8, 5)

(−4, −9)

(−2, −7)

Answer and worked solution

Official answer: D

Differentiate: $2xy^2+2x^2yy'-2=-4y'$ . At $(2,-2)$ : $16-16y'-2=-4y'$ , so $y'=\frac76$ . The tangent is $y+2=\frac76(x-2)$ .

$\left(4,\frac13\right)$ , $(8,5)$ and $(-4,-9)$ lie on it; at $x=-2$ it gives $y=-\frac{20}3\ne-7$ .

JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 71