Question

Let P be a point on the parabola, \( x^2 = 4y \). If the distance of P from the centre of the circle, \( x^2 + y^2 + 6x + 8 = 0 \) is minimum, then the equation of the tangent to the parabola at P, is :

• A. \( x + 4y - 2 = 0 \)
• B. \( x + y + 1 = 0 \)
• C. \( x - y + 3 = 0 \)
• D. \( x + 2y = 0 \)

Hint:

For minimum distance between a curve and a point, the normal at that point must pass through the given point. Finding 't' by inspection in a cubic equation often saves time. \

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The shortest distance between a point on a curve and a fixed point (centre of the circle) occurs along the common normal.

Step 2: Key Formula or Approach:
Circle: \( x^2 + y^2 + 6x + 8 = 0 \implies (x+3)^2 + y^2 = 1 \). Centre is \( C(-3, 0) \).
Parabola: \( x^2 = 4y \). A parametric point on this parabola is \( P(2t, t^2) \).

Step 3: Detailed Explanation:
The slope of the tangent at \( P(2t, t^2) \) to \( x^2 = 4y \) is:
\( 2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2} = \frac{2t}{2} = t \).
The slope of the normal at \( P \) is \( m_n = -\frac{1}{t} \).
The normal passes through centre \( C(-3, 0) \). Equation of normal: \[ y - t^2 = -\frac{1}{t}(x - 2t) \] Substituting \( (-3, 0) \): \[ 0 - t^2 = -\frac{1}{t}(-3 - 2t) \implies t^3 = -3 - 2t \implies t^3 + 2t + 3 = 0 \] By inspection, \( t = -1 \) is a root since \( (-1)^3 + 2(-1) + 3 = 0 \).
So, point \( P \) is \( (2(-1), (-1)^2) = (-2, 1) \).
The equation of the tangent at \( (-2, 1) \) is: \[ x(-2) = 2(y + 1) \implies -2x = 2y + 2 \implies x + y + 1 = 0 \]

Step 4: Final Answer:
The equation of the tangent is \( x + y + 1 = 0 \).