# Tangents and CurvesWatch

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#1
The line x+5y=k is a tangent to the curve x^2 -4y=10. Find the value of the constant k
0
4 weeks ago
#2
(Original post by sneha2002)
The line x+5y=k is a tangent to the curve x^2 -4y=10. Find the value of the constant k
Its similar to a line being tangent to a circle. They cross in 2, 1 or 0 places.
Use the equation of the line to replace one of the variables in the quadratic. Then you have a quadratic and you're looking for the point which has a single solution, ie the discriminant us zero.
Try it and post you're working if there is a problem?
1
#3
I substituted in x=k-5y to get
(k-5y)(k-5y) - 4y =10
k^2 - 10yk +25y^2 - 4y-10=0
But i dont know what to do now

(Original post by mqb2766)
Its similar to a line being tangent to a circle. They cross in 2, 1 or 0 places.
Use the equation of the line to replace one of the variables in the quadratic. Then you have a quadratic and you're looking for the point which has a single solution, ie the discriminant us zero.
Try it and post you're working if there is a problem?
0
4 weeks ago
#4
You have a quadratic in y
You want to find only one solution, so it's where the discriminant is zero, that enables you to find k.
(Original post by sneha2002)
I substituted in x=k-5y to get
(k-5y)(k-5y) - 4y =10
k^2 - 10yk +25y^2 - 4y-10=0
But i dont know what to do now
0
4 weeks ago
#5
wait how do you make the discriminat = 0
0
4 weeks ago
#6
b^2 - 4ac = 0
0
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