Question

Let $S$ be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P ( b, c), b, c \in N$, on the parabola $y^2=2$ ax and the lines $x=b, y=0$ is $16$ unit ${ }^2$, then $\displaystyle\sum_{a \in S} a$ is equal to _____

Hint:

When dealing with tangents and areas, first use the geometric properties of the figure to derive relationships between the variables. Then, use algebraic equations to find the unknowns.

Answer 1

Correct Answer: 146

We are given the parabola: \[ y^2 = 2ax. \] Let the tangent at point \( P(b, c) \) on the parabola. From the equation of the parabola, since \( P(b, c) \) lies on it, we have: \[ c^2 = 2ab \quad \text{(1)}. \] The equation of the tangent to the parabola \( y^2 = 2ax \) at the point \( P(x_1, y_1) = (b, c) \) is: \[ y y_1 = 2a \left( \frac{x + x_1}{2} \right). \] Substituting \( x_1 = b \) and \( y_1 = c \), we get: \[ yc = a(x + b). \] Step 1: For point \( B \), put \( y = 0 \), and now \( x = -b \). Thus, the area of triangle \( \Delta PBA \) is: \[ \text{Area} = \frac{1}{2} \times AB \times AP = 16. \] This simplifies to: \[ \frac{1}{2} \times 2b \times c = 16 \quad \Rightarrow \quad bc = 16. \] Step 2: From equation (1), \( c^2 = 2ab \), so we can solve for \( a \): \[ a = \frac{c^2}{2b}. \] Step 3: Now, possible values of \( (b, c) \) are \( (1, 16), (2, 8), (4, 4), (8, 2), (16, 1) \). 
Step 4: For each pair \( (b, c) \), we can compute \( a \) using the formula \( a = \frac{c^2}{2b} \): - For \( (b, c) = (1, 16) \), \( a = \frac{16^2}{2 \times 1} = 128 \), - For \( (b, c) = (2, 8) \), \( a = \frac{8^2}{2 \times 2} = 16 \), - For \( (b, c) = (4, 4) \), \( a = \frac{4^2}{2 \times 4} = 2 \). Thus, the sum of all possible values of \( a \) is: \[ 128 + 16 + 2 = 146. \]