Conic sections question

Katsura

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I was able to do first part, but the second part just doesn't make much sense. Perhaps it's just a badly structured question. Thanks for help!
 

Otis

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Hello. Let's rename the parabola P because I'm using symbol C for something else, below.

y^2 = 28x

y^2 = (mx + c)^2

If we equate those powers of y, we get a quadratic equation.

If we put that quadratic equation into standard form

Ax^2 + Bx + C = 0​

then we can see expressions for the coefficiants A, B, C.

Edit: We don't need to find the coordinates of the tangent point, but using the quadratic formula to solve Ax^2+Bx+C=0 would give us the x-coordinate. We know a tangent line intersects the given parabola only once, so we want to ensure there's only one solution to Ax^2+Bx+C=0.

Do you remember the discriminant's significance, in the quadratic formula? When the discriminant is zero, there's only one solution for x.


Therefore, substitute the expressions for A,B,C into the discriminant equation

B^2 - 4AC = 0​

and then solve for mc.

Please show any work you can, if you still have questions.

😎
 

MarkFL

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PS: I'm still trying to figure out why this works ...

😎
Equating the discriminant to zero ensures that there is only one root to the equation representing the \(y\)-coordinates of the points of intersection between the line and the parabola. With only 1 point we know the line must therefore be a tangent line.
 

Otis

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Apr 22, 2015
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Equating the discriminant to zero ensures that there is only one root to the equation representing the \(y\)-coordinates of the points of intersection between the line and the parabola. With only 1 point we know the line must therefore be a tangent line.
Thanks, Mark. Yes, I realized that, shortly after logging out. Prior to that, I think I was too focused on potential relationship(s) between the two parabolas. At least my gut instincts are still intact ...
 
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