the tangent at the point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ on the curve $\mathrm{y}=\mathrm{x}^{3}+3 \mathrm{x}^{2}+5$ passes through the origin, then $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ does NOT lie on the curve :

(A) $x^{2}+\frac{y^{2}}{81}=2$

(B) $\frac{y^{2}}{9}-x^{2}=8$

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Application of Derivatives

the tangent at the point

(x_{1}, y_{1})

on the curve

y = x^{3} + 3 x^{2} + 5

passes through the origin, then

(x_{1}, y_{1})

does NOT lie on the curve :

Answer: (D)

The tangent at

(x_{1}, y_{1})

y = x^{3} + 3 x^{2} + 5

y - y_{1} = (3 x_{1}^{2} + 6 x_{1}) (x - x_{1})

. Passing through the origin:

- y_{1} = (3 x_{1}^{2} + 6 x_{1}) (- x_{1})

, so

y_{1} = 3 x_{1}^{3} + 6 x_{1}^{2}

. Since

(x_{1}, y_{1})

is on the curve,

y_{1} = x_{1}^{3} + 3 x_{1}^{2} + 5

. Equating:

3 x_{1}^{3} + 6 x_{1}^{2} = x_{1}^{3} + 3 x_{1}^{2} + 5

, giving

2 x_{1}^{3} + 3 x_{1}^{2} - 5 = 0

. Eliminating

x_{1}

, we find the locus is

2 y = 3 x^{2} + 15

, or

y = \frac{3}{2} x^{2} + \frac{15}{2}

. Checking which curve this does NOT intersect: option (D)

\frac{x}{3} - y^{2} = 2

does not intersect the locus.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option