- Thread starter Katsura
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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.

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

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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 ...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.