find the coordinates of the points on the curve y=x^2+3x+4, the tangents at which passes through the origin - Maths - Application of Derivatives

Question

find the coordinates of the points on the curve y=x^2+3x+4, the
tangents at which passes through the origin

Mayur Pisode · Accepted Answer

Let the required point is x1,y1Now, y = x2 + 3x + 4⇒dydx = 2x + 3⇒dydxx1,y1 = 2x1+ 3Now, equation of tangent at x1,y1 is :y-y1x-x1 = 2x1+ 3Since the tangent passes through the origin 0, 0, so-y1-x1 = 2x1+ 3⇒y1 = 2x12+ 3x1   
1But x1,y1 lies on the given curve, so y1 = x12 + 3x1 + 4   
2Comapring 1 and 2, we get2x12+ 3x1 =  x12 + 3x1 + 4⇒x1 = ±2When x1 = 2; then y1 = 22+32+4 = 4 + 6 + 4 = 14When x1 = -2; then y1 = -22+3-2+4 = 4 - 6 + 4 = 2So, required points are :2,14 and -2,2