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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: nerdmmb on April 11, 2014, 01:45:52 pm

Title: Parabola and Linear Equation Questions
Post by: nerdmmb on April 11, 2014, 01:45:52 pm: I'm a bit confused with finding the value of a pro-numeral in a linear equation that is tangent to a parabola! Would appreciate your help!

First question:
Find the value of c such that y=x+c is a tangent to the parabola y=x²-x-12. (Hint: Consider the discriminant of the resulting quadratic)

Second question:
Find the values of m for which the straight line y=mx+6 is tangent to the parabola y=-2x²-6x+2. (Hint: Use the discriminant of the resulting quadratic)

I also don't understand why the question has hinted us to use the discriminant.
Title: Re: Parabola and Linear Equation Questions
Post by: IndefatigableLover on April 11, 2014, 02:14:07 pm: Quote from: nerdmmb on April 11, 2014, 01:45:52 pm
I'm a bit confused in finding the value of a pro-numeral in a linear equation that is tangent to a parabola! Would appreciate your help!

First question:
Find the value of c such that y=x+c is a tangent to the parabola y=x²-x-12. (Hint: Consider the discriminant of the resulting quadratic)

Second question:
Find the values of m for which the straight line y=mx+6 is tangent to the parabola y=-2x²-6x+2. (Hint: Use the discriminant of the resulting quadratic)

I also don't understand why the question has hinted us to use the discriminant.
If you that it is a tangent then it will only touch it once. In this case you will have only one solution and when using the discriminant, this will equal zero.

So going with the first question, if we know that y = x+c is a tangent of the quadratic then we can equate so it's:

$x^2-x+12 = x+c$ (Drag everything to one side)

$x^2-2x-(12+c)=0$ (We can put 12+c together since they go together anyway).

Now normally we could try and quadratic solve but that's not really going to help us in this case because we have this unknown pronumeral. What we can do is use the discriminant as we only worry about the coefficients of 'x' or whatever and we can then just transpose for 'c' since it'll be the only pronumeral left after we use it.

You know the formula for the discriminant is just b^2-4ac (where a = 1, b= -2, c= (12+c))

$\therefore \triangle = 4+4(12+c)$ (expand)

$= 4c+52 = 0$ (move 52 and divide by 4)

$\therefore c=-13$

2. Same thing as before. We equate the two equations together and make the discriminant equal to zero since it's just a tangent (or only one solution).

$-2x^2-6x+2 = mx+6$

$\therefore -2x^2-(6+m)x-4$ (I put the 'x' at the back because (6+m) is the coefficient in this case or what we'll be using for the discriminant so not to be confused with)

$\therefore 2x^2+(6+m)x+4=0$ (take out the negative)

Use the discriminant rule

$\therefore \triangle = (6+m)^2-4(4)(2)=0$

$\therefore (6+m)^2=32$ (put 32 on the other side)

$6+m = \pm \sqrt{32}$

$\therefore m = -6 \pm \sqrt{32}$

Tidying it up at the end, you can rewrite the square root 32 into a surd because 16 goes into it perfectly.

$\therefore m=-6 \pm 4\sqrt{2}$
Title: Re: Parabola and Linear Equation Questions
Post by: nerdmmb on April 11, 2014, 02:57:42 pm: Quote from: IndefatigableLover on April 11, 2014, 02:14:07 pm
If you that it is a tangent then it will only touch it once. In this case you will have only one solution and when using the discriminant, this will equal zero.

So going with the first question, if we know that y = x+c is a tangent of the quadratic then we can equate so it's:

$x^2-x+12 = x+c$ (Drag everything to one side)

$x^2-2x-(12+c)=0$ (We can put 12+c together since they go together anyway).

Now normally we could try and quadratic solve but that's not really going to help us in this case because we have this unknown pronumeral. What we can do is use the discriminant as we only worry about the coefficients of 'x' or whatever and we can then just transpose for 'c' since it'll be the only pronumeral left after we use it.

You know the formula for the discriminant is just b^2-4ac (where a = 1, b= -2, c= (12+c))

$\therefore \triangle = 4+4(12+c)$ (expand)

$= 4c+52 = 0$ (move 52 and divide by 4)

$\therefore c=-13$

2. Same thing as before. We equate the two equations together and make the discriminant equal to zero since it's just a tangent (or only one solution).

$-2x^2-6x+2 = mx+6$

$\therefore -2x^2-(6+m)x-4$ (I put the 'x' at the back because (6+m) is the coefficient in this case or what we'll be using for the discriminant so not to be confused with)

$\therefore 2x^2+(6+m)x+4=0$ (take out the negative)

Use the discriminant rule

$\therefore \triangle = (6+m)^2-4(4)(2)=0$

$\therefore (6+m)^2=32$ (put 32 on the other side)

$6+m = \pm \sqrt{32}$

$\therefore m = -6 \pm \sqrt{32}$

Tidying it up at the end, you can rewrite the square root 32 into a surd because 16 goes into it perfectly.

$\therefore m=-6 \pm 4\sqrt{2}$

Thanks heaps IndefatigableLover!
Title: Re: Parabola and Linear Equation Questions
Post by: nerdmmb on April 11, 2014, 04:16:52 pm: Also had another question:
Find the equation of the straight line(s) which pass through the point (1,-2) and is (are) tangent to the parabola with equation y=x²

Thanks :D
Title: Re: Parabola and Linear Equation Questions
Post by: IndefatigableLover on April 11, 2014, 05:51:30 pm: Quote from: nerdmmb on April 11, 2014, 04:16:52 pm
Also had another question:
Find the equation of the straight line(s) which pass through the point (1,-2) and is (are) tangent to the parabola with equation y=x²

Thanks :D
Not sure if you know the formula but when you have a point and it is a tangent then you can use the formula:

$y-y_1=m(x-x_1)$ (where (x1,y1) are points). (This also leads onto tangents and normals but we'll leave that for today because the question doesn't require it).

Sub in the point (1,-2) into the formula

$y-(-2)=m(x-(1))$

$y=m(x-1)-2$

Because it's a tangent of the parabola then we can equate the two equations together,

$x^2 = m(x-1)-2$ (And transpose)

$x^2-m(x-1)+2=0$ (Expand)

$x^2-mx+(m+2)=0$

Chuck in the discriminant

$\therefore \triangle = m^2-(4)(1)(m+2)$

$m^2-4m-8=0$

Okay so we can't really solve it straight away like we did with the other two questions... normally we could factor this and find what 'm' is but we can't so the easiest route is to just use the quadratic formula and solve for 'm' really.

In a rush (Latex takes a while) but you can use the quadratic formula for that
Spoiler
And you should get $m=2\pm \sqrt{12}$ (you can make that square root into a surd as well)

Afterwards you just plug that value back into $y=m(x-1)-2$ and you've got your two answers! It would be best once you sub those in that you expand it out (since it looks kind of ugly if you don't expand it I guess but should still be alright I dunno...)

If you need more help for the answer then I can guide you through it more if you want :)
Title: Re: Parabola and Linear Equation Questions
Post by: nerdmmb on April 11, 2014, 09:21:27 pm: @IndefatigableLover

THANKYOU THANKYOU THANKYOU!! :D