Find the coordinate of point A in the figure Can you solve it in way which doesn't involve derivation?
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Marva Jami
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- Compute the equation of the blue line. 2. Compute the coordinate of the second intersection point $P$ between the blue line and the red parabola. 3. Find an equation for the line (PA). 4. Conclude.
– Clément Guérin Feb 10 '18 at 07:08 -
I'm not sure how I'd solve it with derivation. Just solve equations. The y-intercept is -2, so the blue line is y = -x -2. So the blue line intercepts the parabola when y= -x-2 =-x^2 + 4 so x^2-x -6=0 or x=3 and y=-5. so the equation of the perpendicular line is $y+5 = x-3$ then so A is where $0+5 = x -3$ or $(x,y) = (8,0)$. – fleablood Feb 10 '18 at 07:23
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The parabola has zeros at $x=\pm 2$. Your blue line passes through $(-2,0)$ and $(0,-2)$ so its equation is $y=-x-2$. The blue line intersects the parabola a second time when $-(x+2) = -(x-2)(x+2)$ with $x\neq -2$, which occurs at $(3,-5)$. The black line is perpendicular to the blue one so it must have slope $1$. Therefore, its equation is $y = x-8$. So the point $A$, which is the $x-$intercept of the black line, must be $(8,0)$.
bames
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Do you know the formula $y-y_0 = m(x-x_0)$? We know the black line passes through $(3,-5)$, so just plug in $m=1$ and $(x_0,y_0) = (3,-5)$ to obtain $y-(-5) = x-3 \longrightarrow y = x - 8$. – bames Feb 10 '18 at 07:30
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Another approach: The coordinates of intersections between the parabola and the straight line are $(-2,0)$ and $(3,-5)$. Now apply the geometric mean theorem on the triangle formed by those two points and $A$.
Michael Hoppe
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