The area of the triangle formed by the positi | vedclass

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(A) The equation of the circle is $x^2 + y^2 = 4$. The point of contact is $P(1, \sqrt{3})$.The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_

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$2\sqrt{3}$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 = 4$. The point of contact is $P(1, \sqrt{3})$.The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$,so $x(1) + y(\sqrt{3}) = 4$,which is $x + \sqrt{3}y = 4$.For the tangent,the $x$-intercept is found by setting $y=0$,giving $x=4$. So the tangent meets the $x$-axis at $A(4, 0)$.The normal at any point on the circle passes through the center $(0, 0)$. The normal at $(1, \sqrt{3})$ is the line passing through $(0, 0)$ and $(1, \sqrt{3})$,which is $y = \sqrt{3}x$.The normal meets the $x$-axis at the origin $O(0, 0)$.The triangle is formed by the points $O(0, 0)$,$A(4, 0)$,and $P(1, \sqrt{3})$.The base of the triangle along the $x$-axis is the distance $OA = 4$.The height of the triangle is the $y$-coordinate of $P$,which is $\sqrt{3}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes \sqrt{3} = 2\sqrt{3}$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

The area of the triangle formed by the positive $x$-axis and the normal and the tangent to the circle $x^2 + y^2 = 4$ at $(1, \sqrt{3})$ is

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Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
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$5. -2\sqrt{2}$

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