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The line x+5y=k is a tangent to the curve x^2-4y=10 . Find the value of the constant k.

Have done k^2-10yk+25y^2-4y-10=0 by inputting x+5y=k into x^2-4y=10 but unsure what to do next?
Reply 1
As its a tangent, b^2 - 4ac will equal 0.
A= 25, b=4+10k, c=k^2 - 10
Original post by Nonagone
As its a tangent, b^2 - 4ac will equal 0.
A= 25, b=4+10k, c=k^2 - 10

Thanks but i dont understand why would a, b and c equal that?
Reply 3
Original post by Username123455
Thanks but i dont understand why would a, b and c equal that?

Thats because when you get that equation, you get an equation in terms of y^2, y's and constants. Which is, 25y^2, - (4+10k)y - 10 + k^2 = 0.
The way we could solve this normally (if we knew k) , would be the quadratic formula. As you hvae it in terms of y, this would give us our solutions for the y values.
For example if we tried to solve x^2 + 10x - 2 = 0, we would say a=1, b=10 and c=-2.
The equation we are using represents the two lines crossing.
We need it so that if we solved the equation, for example by using the quadratic formula, we get one solution (as this is how many values we would expect if it was a tangent as it would only touch the line once, giving one x value (or y value in this case)).
( If it touched twice this would give a lot of values for k )
The only way we can get one solution from the quadratic formula is if the part in the square root is 0. This is because if it was positive, we would always get two values, and if its negative we are working with imaginary numbers.
The part in the square root is b^2 - 4aC. this would therefore need to be equal to 0 to give us one solution.
Using this, we can then get a new equation in terms of k. We can solve it to figure out the answer.
Original post by Nonagone
Thats because when you get that equation, you get an equation in terms of y^2, y's and constants. Which is, 25y^2, - (4+10k)y - 10 + k^2 = 0.
The way we could solve this normally (if we knew k) , would be the quadratic formula. As you hvae it in terms of y, this would give us our solutions for the y values.
For example if we tried to solve x^2 + 10x - 2 = 0, we would say a=1, b=10 and c=-2.
The equation we are using represents the two lines crossing.
We need it so that if we solved the equation, for example by using the quadratic formula, we get one solution (as this is how many values we would expect if it was a tangent as it would only touch the line once, giving one x value (or y value in this case)).
( If it touched twice this would give a lot of values for k )
The only way we can get one solution from the quadratic formula is if the part in the square root is 0. This is because if it was positive, we would always get two values, and if its negative we are working with imaginary numbers.
The part in the square root is b^2 - 4aC. this would therefore need to be equal to 0 to give us one solution.
Using this, we can then get a new equation in terms of k. We can solve it to figure out the answer.

Thnak you!
You could follow the other persons method, or another simple method is finding the gradient of the tangent is -1/5, so first find the derivative of the curve which is x/2 so find value of x when equation of range is -1/5 so x/2=-1/5 so x=-2/5, then sub in -2/5 into the equation of curve to get value of y= -2.46, now sub in x and y into equation of tangent to get value of k to be= -12.7

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