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Question 4
In the diagram below, the circle with centre S, passes through the origin, O, and intersects the x-axis at R and y-axis at T. The tangent to the circle at P(4 ; -6) ... show full transcript
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Answer
To find the equation of the tangent line at P(4, -6), we first find the slope (m) of the radius SP:
(m_{SP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - (-3)}{4 - 2} = \frac{-3}{2}.
The slope of the tangent line is the negative reciprocal, so:
(m_{UQ} = \frac{2}{3}.
Using point-slope form (y - y_1 = m(x - x_1)), with point P(4, -6):
(y + 6 = \frac{2}{3}(x - 4),)
which simplifies to:
(y = \frac{2}{3}x - \frac{8}{3} - 6 = \frac{2}{3}x - \frac{26}{3}.)
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Answer
To find this ratio, we first calculate the areas. Area AOTP is a triangle with vertices at A(0, 0), O(2, -3), T(0, -6).
Using the formula for the area of a triangle:
,
we can find the area of both triangles:
Area AOTP = (\frac{1}{2}\times 2 \times 6 = 6.
Area APTU = 8.
Thus, the ratio is (\frac{6}{8} = \frac{3}{4}.)
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