The correct option is D (-2, 2)
Let P(h, k) be a point on the given curve such that the tangent at P passes through the origin. Since, P(h, k) lies on the given curve y=x2+3x+4.
∴k=h2+3h+4 ....[1]
The equation of the curve is y=x2+3x+4
Differentiating w.r.t. x, we get
dydx=2x+3⇒(dydx)P=2h+3.
The equation of the tangent at P(h, k) is
y−k=(dydx)P(x−h)⇒y−k=(2h+3)(x−h)It passes through the origin i.e. (0, 0).∴0−k=(2h+3)(0−h)⇒k=2h2+3h ...(2]Subtracting (2] from (1], we get−h2+4=0⇒h=±2From (2],If h=2, then y=14If h=−2, then y=2
Hence, the required points are (2, 14) and (-2, 2).