The curve y = 10 /2x + 1 − 2 intersects the x-axis at A. The tangent to the curve at A intersects the y-axis at C. (i) Show that the equation of AC is 5y + 4x = 8. [5] (ii) Find the distance AC.
kashol123:
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y=10/(2x+1)-2 intersect the x axis hence
y=0 so, 0=10/(2x+1)-2
5=2x+1
x=2
so point A (2,0)
now
differentiate y=10/(2x+1)-2 w.r.t x
dy/dx= -10/(2x+1)^2.2=-20/(2x+1)^2
slope at ( 2,0)=-20/(5)^2=-4/5
now equation of line AC
(y-0)=-4/5 (x-2)
5y+4x=8
now point C find
because C is at y axis so x=0
put this in equation then y =8/5
so C (0,8/5)
now distance b/w AC=root {(2)^2+(8/5)^2}
y=0 so, 0=10/(2x+1)-2
5=2x+1
x=2
so point A (2,0)
now
differentiate y=10/(2x+1)-2 w.r.t x
dy/dx= -10/(2x+1)^2.2=-20/(2x+1)^2
slope at ( 2,0)=-20/(5)^2=-4/5
now equation of line AC
(y-0)=-4/5 (x-2)
5y+4x=8
now point C find
because C is at y axis so x=0
put this in equation then y =8/5
so C (0,8/5)
now distance b/w AC=root {(2)^2+(8/5)^2}
I hope you understand
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please reply
when you differentiated it why did you times it by 2
actually 1/x=x^-1 you written
now x^-1 differentiation -x^-2 i.e -1/x^2 in the same way I did it
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