Question 528215
can you please help me solve these problems on parabolas and circles? Thanks
1.Find the equation of the two tangents to the parabola y^2=5x from the point (-1,1).
2.Find the equation of the two tangents,drawn through the external points (11,3) to the circle x^2+y^2=40.
3.write the equations of the tangent to the parabola y^2=10x,at the extremities of the latus recum.on what line do these tangent intersect.
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1.Find the equation of the two tangents to the parabola y^2=5x from the point (-1,1).
y^2=5x
x=y^2/5
point slope formula: y-y1=m(x-x1)
For given problem:
y-1=m(x+1)
y-1=mx+m
y-1=m((y^2)/5)+1)
y-1=(m/5)y^2+m
(m/5)y^2+m-y+1=0
a=m/5, b=-1, c=(m+1)
b^2-4ac=1-4*(m/5)(m+1)
=1-(4/5)m^2-(4/5)m
=.8m^2+.8m-1=0
a=.8, b=.8, c=-1
m=[-.8±√(.8^2-4*.8*-1)]/2*.8
m=[-.8±√3.84]/1.6
m=(-.8±1.96)/1.6
m=-1.72
or
m=.724
Equation of line:
y=mx+b
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y=-1.72x+b
solve for b using given point (-1,1)
1=-1.72*-1+b
b=-.72
equation: y=-1.72x-.72
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y=.724x+b
solve for b using given point (-1,1)
1=.724*-1+b
b=1.724
equation: y=.724x+1.724
see graph below as a visual check on the answers
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y=±(5x)^.5

 {{{ graph( 300, 300, -10, 10, -10, 10,(5x)^.5,-(5x)^.5,-1.72x-.72,.724x+1.724) }}}

note to student: Please submit other two problems separately. 
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