Question

Find the shortest distance between the line y = x – 2 and the curve y = x² + 3x + 2

Answer 1

L₁ || L₂

\(m_{L_1}=m_{L_2}\)

\(m_{L_1}=1\)

L₁ is tangent to Parabola

\(y = x^2+3x+2\)

\(\frac{dy}{dx}=2x+3\)

\((\frac{dy}{dx})_{a,b}=2a+3=1\)

2a = -2

a = -1

b = a² + 3a + 2

b = (-1)² + 3(-1) + 2

b = 3 - 3 

b = 0

P(a,b) ⇒ P(-1,0)

ax + by + c = 0  (h,k)

\(d=|\frac{ah+bk+c}{\sqrt{a^2+b^2}}|\)

x - y - 2 = 0  (-1, 0)

\(d=|\frac{-1-0-2}{\sqrt{(1)^2}+(-1)^2}|\)

\(=|\frac{-3}{\sqrt{2}}|\)

\(=\frac{3}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\)

\(d=\frac{3\sqrt{2}}{2}\)

Answer 2

The given curves are y = x² + 3x + 2 and y = x – 2

⇒ dy/dx = 2x + 3 and dy/dx = 1

Since the tangents are parallel, so their slopes are same. Therefore, 2x + 3 = 1

⇒ x = -1 when x = 1, then y is 6.

Thus, the point lies on the curve is (1, 6)

Hence, the length of the shortest distance