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 _____

Answer 1

Correct Answer: 146

To solve the problem, we first consider the parabola \(y^2 = 2ax\). The equation of the tangent to the parabola at point \(P(b, c)\) can be derived. Since \(P\) lies on the parabola, \(c^2 = 2ab\). The standard form of the tangent at any point \((x_1, y_1)\) on the parabola \(y^2 = 2ax\) is \(yy_1 = a(x + x_1)\). Substituting \(P(b, c)\) gives the tangent as \(yc = a(x + b)\) or \(yc = ax + ab\).
Next, we find the area of the triangle formed by this tangent, the vertical line \(x = b\), and the x-axis \(y = 0\). The x-intercept of the tangent is found by setting \(y = 0\):
\[ c \times 0 = ax + ab \implies ax + ab = 0 \implies x = -b \text{ (since \(a \neq 0\))}\]
Thus, the x-intercepts are \(x = -b\) and \(x = b\), and the y-intercept is at \(y = 0\). Therefore, the vertices of the triangle are \((-b, 0), (b, 0), \text{and } (0, c)\).
The base of the triangle is the distance between \((-b, 0)\) and \(b, 0\), which is \(2b\). The height is \(c\). The area of the triangle is:
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2b \times c = bc\]
We are given that the area is 16, hence:
\[bc = 16\]
Substitute \(c = \frac{16}{b}\) into the parabola equation \(c^2 = 2ab\):
\[\left(\frac{16}{b}\right)^2 = 2ab \implies \frac{256}{b^2} = 2ab \implies 256 = 2ab^3\]
Solving for \(a\) in terms of \(b\) gives:
\[a = \frac{256}{2b^3} = \frac{128}{b^3}\]
Since \(a\) is a natural number, \(\frac{128}{b^3} \in \mathbb{N}\) implies \(b^3\) must divide 128. The integer divisors of 128 are calculated as \(1, 2, 4, 8, 16, 32, 64, 128\). Therefore, the possible values for \(b\) such that \(b^3\) divides 128 are \(b = 1, 2, 4\), because:
- \(1^3 = 1\), \(2^3 = 8\), \(4^3 = 64\), all divide 128.
Calculate corresponding \(a\):
- For \(b = 1\), \(a = \frac{128}{1} = 128\)
- For \(b = 2\), \(a = \frac{128}{8} = 16\)
- For \(b = 4\), \(a = \frac{128}{64} = 2\)
Thus, the set \(S = \{128, 16, 2\}\). The sum \(\sum_{a \in S} a = 128 + 16 + 2 = 146\), which is within the range (146,146) as expected.