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Maths help !

The line x+5y=k is a tangent to the curve x^2-4y=10. Find the value of the constant k. I get to k^2+25y^2-10ky-4y-10=0 and I know that I need to use the discriminant but I don't know how to get there.
Reply 1
Avatar for Pangol
Pangol
15
Original post by bongamin12
The line x+5y=k is a tangent to the curve x^2-4y=10. Find the value of the constant k. I get to k^2+25y^2-10ky-4y-10=0 and I know that I need to use the discriminant but I don't know how to get there.

If I were you I would eliminate y to get a quadratic in x - it will be easier to work with. But it will work either way.

If the line is a tangent to the parabola, how many points are there in common between the line and the curve? What does this mean for the number of solutions that your equation will have? Hopefully you can see how the discriminant comes into it now.
Here's some hints:

The discriminant tells you how many real solutions to a quadratic equation there will be.

Think about if the line is a tangent to a curve (try drawing it out if it helps), at how many points does the tangent intercept with the curve?

Original post by bongamin12
The line x+5y=k is a tangent to the curve x^2-4y=10. Find the value of the constant k. I get to k^2+25y^2-10ky-4y-10=0 and I know that I need to use the discriminant but I don't know how to get there.
Reply 3
Avatar for bongamin12
bongamin12
OP
2
I am not sure how to get the values out for the discriminant with the equation I currently have tho.
Reply 4
Avatar for vbzl
vbzl
9
Original post by bongamin12
I am not sure how to get the values out for the discriminant with the equation I currently have tho.

Can you show your work so far?
(edited 4 years ago)
Reply 5
Avatar for simon0
simon0
13
You stated in your first post you had equation: k2+25y2−10ky−4y−10=0. k^{2} + 25 y^{2} - 10ky - 4y -10 = 0.

As others have stated, as the tangent meets the curve once the discriminant (of the quadratic) has the case: b2=4ac. b^{2} = 4ac.
We can use this to obtain the y coordinate where the line meets the curve.

Rearranging the above quadratic for powers of y and using the discriminant case, can you carry on from here?
(edited 4 years ago)
Reply 6
Avatar for simon0
simon0
13
Otherwise you can:
1) Use @Pangol 's advice in elmininating the y variable. This saves on the amount of calculations and needing to square.

2) Use differentiation as we know the gradient of the line and dy/dx of the curve can be calculated.
Then we can easily obtain the coordinates of where the line meets the curve and use this to find k.

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