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How to find this tangent?

Okay so I have a parabola with equation y2=xy^2=x and I need to find the tangent to the parabola at the point (16,−4)(16,-4) I tried setting y2=(mx+c)2=0y^2=(mx+c)^2=0 and then trying to get mm using that b2−4ac=0b^2-4ac=0 since it's a tangent but that didn't work and I don't really know what to do now.

Any advice?

I will check back in ∼\sim 1hr.
(edited 9 years ago)
Reply 1
Original post by poorform
Okay so I have a parabola with equation y2=xy^2=x and I need to find the tangent to the parabola at the point (16,−4)(16,-4) I tried setting y2=(mx+c)2=0y^2=(mx+c)^2=0 and then trying to get mm using that b2−4ac=0b^2-4ac=0 since it's a tangent but that didn't work and I don't really know what to do now.

Any advice?

I will check back in ∼\sim 1hr.


Why have you set that expression equal to 0?

Just substitute y = mx + c into the original and solve.

Alternatively, work out the gradient of the parabola at the given point and use that to work out the equation of the tangent
Original post by poorform
Okay so I have a parabola with equation y2=xy^2=x and I need to find the tangent to the parabola at the point (16,−4)(16,-4) I tried setting y2=(mx+c)2=0y^2=(mx+c)^2=0 and then trying to get mm using that b2−4ac=0b^2-4ac=0 since it's a tangent but that didn't work and I don't really know what to do now.

Any advice?

I will check back in ∼\sim 1hr.


My initial thoughts would be to differentiate y2=xy^2=x and rearrange to get dydx=...\frac{dy}{dx}=.... Plug in numbers to get gradient of the tangent at that point i.e. m. Then you can use y−y1=m(x−x1)y-y_1 = m(x-x_1), where (x1,y1)(x_1, y_1) is the point they've given you.
Reply 3
Original post by davros
Why have you set that expression equal to 0?

Just substitute y = mx + c into the original and solve.

Alternatively, work out the gradient of the parabola at the given point and use that to work out the equation of the tangent


I'm fine using the second method but I'd prefer not to use any calculus at this point. I still can't figure out what to do could you look through what I've done and help.

y2=x\displaystyle y^2=x

y=mx+c\displaystyle y=mx+c

y2=(mx+c)2=m2x2+2mxc+c2−x\displaystyle y^2=(mx+c)^2=m^2x^2+2mxc+c^2-x the line is a tangent   ⟹  \displaystyle \implies discriminant must be equal to zero and so (2xc)2−4x2(c2−x)=0  ⟺  4x2c2−4x2c2−4x3=0  ⟺  −4x3=0\displaystyle (2xc)^2-4x^2(c^2-x)=0 \iff 4x^2c^2-4x^2c^2-4x^3=0 \iff -4x^3=0.

Have I done something wrong I'm not too sure I understand exactly what it going on here so if you could help me I would appreciate it. Thanks.
Reply 4
Original post by poorform
I'm fine using the second method but I'd prefer not to use any calculus at this point. I still can't figure out what to do could you look through what I've done and help.

y2=x\displaystyle y^2=x

y=mx+c\displaystyle y=mx+c

y2=(mx+c)2=m2x2+2mxc+c2−x\displaystyle y^2=(mx+c)^2=m^2x^2+2mxc+c^2-x the line is a tangent   ⟹  \displaystyle \implies discriminant must be equal to zero and so (2xc)2−4x2(c2−x)=0  ⟺  4x2c2−4x2c2−4x3=0  ⟺  −4x3=0\displaystyle (2xc)^2-4x^2(c^2-x)=0 \iff 4x^2c^2-4x^2c^2-4x^3=0 \iff -4x^3=0.

Have I done something wrong I'm not too sure I understand exactly what it going on here so if you could help me I would appreciate it. Thanks.


Anyone?
Reply 5
Original post by poorform
Anyone?


Nevermind got it.
Reply 6
Original post by poorform
Anyone?


x shouldn't be part of the discriminant since you have a quadratic in the variable x!

Original post by poorform
Nevermind got it.


Good stuff :smile:

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