Is there an intersection between the line (Y = aX - b) and the parabol

So here is my first attempt at posting a reply to any question. Help me if i go wrong

To figure out whether these two intersect, we can solve their equations.
Equating the RHS of both equations, we get :
\(ax - b = x^2 + b\)
\(x^2 - ax + 2b = 0\)

This is of the form\(A x^2 + B x + C = 0\)
So if the equation has real numbers as roots, we can conclude that the parabola and circle do intersect.

For the equation to have real roots, \(B^2 - 4AC\) should be a positive number, so that \(\sqrt{B^2-4AC}\) is real.

In our equation \(B^2 - 4AC = a^2 - 4*1*2b = a^2 - 8b\)
We don't know whether \(a^2> 8b\) from (1). Therefore we don't know if the equation has real roots.
From (2), we can conclude that since b<0 , -8b>0. Therefore, \(B^2-4AC\) is positive. Hence, the answer should be B.

Therefore, in my opinion the answer should be E.

Apoorva, Why did you rule out Statement2?

Statement2 says that b<0, which means it is negative. And, \(a^2\) is definitely a non negative value.
Which would mean that \(a^2 - 8b\) is a positive value and the equation in question can have real roots.

Yep totally missed that. Thank you

i thought this way.

since the min point of this parabola is (0,b) thus it is symmetrical around the y axis and since the co-oficient of x^2 is 1 then it is opened upwards and its x axis intercepts are dependent on b (+or- (-4b)^1/2. if b is -ve then it has 2 x intercepts and y intercept is -ve and and if +ve then it has no x intercept and y intercept is +ve.

in all cases ( regardless of slope) if the line y-intercept is > the y intercept of the parabola it ll always intersect it thus the solution to this question depends on (b)

from one no idea about b ...insuff

from 2 , b is -ve thus the y intercept of the parabola is -ve and that of the line is +ve thus they surely intersect.

B

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