Let be the set of all <img alt="\mathrm{a\in N}" src="https://entrancecorner.oncodecogs.com

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

Option: 1
146

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

tangent at $\mathrm{P(b, c) my^{2} = 2ax}$ is

$\mathrm{ \text { area }=\left|\frac{1}{2} \times 2 \mathrm{~b} \times \frac{2 \mathrm{ba}}{\mathrm{c}}\right|=16 }$
$\mathrm{ \frac{2 \mathrm{~b}^2 \mathrm{a}}{\mathrm{C}}=16 }$
$\mathrm{ \frac{\mathrm{b}^2 \mathrm{a}}{\mathrm{C}}=8 }$
$\mathrm{ \because \mathrm{P}(\mathrm{b}, \mathrm{c}) \text { lies on } \mathrm{y}^2=2 \mathrm{ax} }$
$\mathrm{ \mathrm{C}^2=2 \mathrm{ab} }$
$\mathrm{ \Rightarrow \quad \frac{\mathrm{b}^4 \mathrm{a}^2}{\mathrm{c}^2}=64 }$
$\mathrm{ \Rightarrow \quad \frac{\mathrm{b}^4 \mathrm{a}^2}{2 \mathrm{ab}}=64 }$
$\mathrm{ \Rightarrow \quad \mathrm{b}^3 \mathrm{a}=128 }$
$\mathrm{ \Rightarrow \quad a=\frac{128}{\mathrm{~b}^3} }$
$\mathrm{ \mathrm{a} \: can\: \mathrm{be} \: 128,16,2 \text { then } }$
$\mathrm{ \mathrm{S}=\{2,16,128\} }$
$\mathrm{ \sum \mathrm{a}=146}$