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# Find the area of triangle by tangents at the point $P(at_1^2,2at_1)$ & $Q(at_2^2,2at_2)$ for the parabola $y^2=4ax$ .

$\begin{array}{1 1}(a)\;\large\frac{1}{2}\normalsize \mid a^2(t_1-t_2)^3\mid\\(b)\;\large\frac{1}{4}\normalsize\mid a^2(t_1-t_2)^3\mid\\(c)\;\large\frac{1}{2}\normalsize\mid a^2(t_1+t_2)^3\mid\\(d)\;\large\frac{1}{2}\normalsize\mid a^2(t_2-t_1)^3\mid\end{array}$

Area of $\Delta PAQ=\large\frac{1}{2}$$\begin{vmatrix}at_1^2&2at_1&1\\at_2^2&2at_2&1\\at_1t_2&a(t_1+t_2)&1\end{vmatrix} Solving the determinant \Rightarrow \large\frac{1}{2}$$\mid(at_1^2(2at_2-a(t_1+t_2))-2at_1(at_2^2-at_1t_2))+(1(at_2^2\times a(t_1+t_2)-at_1t_2\times 2at_2))\mid$
$\Rightarrow \large\frac{1}{2}$$\mid (a^2t_1^2t_2-a^2t_1^3-2a^2t_1t_2^2+2a^2t_1^2t_2+a^2t_2^3-a^2t_2^2t_1)\mid \Rightarrow \large\frac{1}{2}$$\mid (3a^2t_1^2t_2-3a^2t_2^2t_1+a^2t_2^3-a^2t_1^3)\mid$
$\Rightarrow \large\frac{1}{2}$$\mid a^2(3t_1^2t_2-3t_2^2t_1+t_2^3-t_1^3)\mid \Rightarrow \large\frac{1}{2}$$\mid (a^2(t_1-t_2)^3\mid$
Hence (a) is the correct answer.