Let A(1, 2),B(4, -4),and C(2, 2sqrt{2}) be po | vedclass

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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\alpha = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\alpha = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. \ Substituting the coordinates $A(1, 2), B(4, -4), C(2, 2\sqrt{2})$: \ $\alpha = \frac{1}{2} |1(-4 - 2\sqrt{2}) + 4(2\sqrt{2} - 2) + 2(2 - (-4))|$ \ $\alpha = \frac{1}{2} |-4 - 2\sqrt{2} + 8\sqrt{2} - 8 + 12| = \frac{1}{2} |6\sqrt{2}| = 3\sqrt{2}$. \ For a parabola $y^2 = 4ax$,the area of the triangle formed by the tangents at three points is half the area of the triangle formed by the points themselves. \ Thus,$\alpha = 2\beta$,which implies $\beta = \frac{\alpha}{2} = \frac{3\sqrt{2}}{2}$. \ Therefore,$\alpha \beta = (3\sqrt{2}) 	imes \left(\frac{3\sqrt{2}}{2}ight) = \frac{9 	imes 2}{2} = 9$.

Let $A(1, 2)$,$B(4, -4)$,and $C(2, 2\sqrt{2})$ be points on the parabola $y^2 = 4x$. If $\alpha$ and $\beta$ respectively represent the area of $\triangle ABC$ and the area of the triangle formed by the tangents at $A, B, C$ to the parabola,then $\alpha \beta =$

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