A line L : y = m x + 3 meets y - a x is at E(0,3) and the arc of the parabola y 2 = 1 6 x , 0 ≤ y…

1. a-p;b-q;c-s;d-r;
2. a-s;b-p;c-q;d-r;
3. a-r;b-q;c-s;d-p;
4. a-p;b-q;c-s;d-r;

Solution

tangent at F      yt = c + 4t²
a : x = 0        y = 4t   (0,4t)
(4t²,8t) satisfies the line
8t = 4mt² + 3
4mt² - 8t + 3 = 0
$\text{Area}=\frac{1}{2}\left|\begin{array}{ccc}0& 3& 1\\ 0& 4t& 1\\ 4t^2& 8t& 1\end{array}\right|$

$=\frac{1}{2}\left(4t^2\left(3-4t\right)\right)$
$=2t^2\left(3-4t\right)$
$\text{A}=2\left[3t^2-4t^3\right]$
$=\frac{\text{dA}}{\text{dt}}=2\left[6t-12t^2\right]$
= 24 t(1 - 2t)
​
t = 1/2 maxima
$\text{G}\left(0,4t\right)\Rightarrow \text{G}\left(0,2\right)$
y₁ = 2
$\left(x_0,y_0\right)=\left(4t^2,8t\right)=\left(1,4\right)$
$y_0=4$
$\text{Area}=2\left(\frac{3}{4}-\frac{1}{2}\right)=2\left(\frac{3-2}{4}\right)=\frac{1}{2}$

Asked in: JEE Advanced 2013 (Paper 2)

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