The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two

yashikaaggarwal


\(PO=x_2 - x_1 = c\)
\(QO = y_2-y_1 = 3c\)

\(PO^2+QO^2 = c^2+9c^2 = PQ^2\)

c= +-4

Since \(x_2>x_1\)

\(x_2 = x_1+4 \)

I just assumed x_1=t and x_2 = t+4 in my previous solution.

As, Line and parabola intersect at t and t+4, both values must satisfies \(x^2 + ax + 4 = 3x+b\) or are the roots of this equation.


Sum of the roots of \(x^2+(a-3)x+4-b = 0\) is -(a-3)
t+t+4 = 3-a

If you still have any doubt, you can ask.

Just last question why did you took slope 3 as the difference. Won't constant play any role?

Posted from my mobile device

constant term won't affect the slope of the line. Change in constant term just shifts the line upwards or downwards on coordinate plane.

y=3x+b; slope is 3 and y-intercept of the line = b


Slope of the line\(=3 = \frac{y_2-y_1}{x_2-x_1} \)

\(3 = \frac{y_2-y_1}{c}\)

\(y_2-y_1 = 3c \)

Got it. Thank you for being patient.

Given:
1. The intersections of the graph \(y = x^2 + ax + 4\) and \(y = 3x + b\) are two points \(P\) and \(Q\).
2. The length of \(PQ\) is \(4\sqrt{10}.\)

Asked: What is the maximum of \(b\)?

\(y = x^2 + ax + 4 = 3x + b\)
\(x^2 + (a-3)x + (4-b) = 0\)

\(x_1 + x_2 = (3-a)\)
\(x_1*x_2 = (4-b)\)

\((x_1 - x_2)^2 = (3-a)^2 - 4(4-b) = 9 + a^2 - 6a - 16 + 4b = a^2 - 6a + 4b - 7\)

\((y_1 - y_2)^2 = (3(x_1-x_2))^2 = 9(a^2 - 6a - 7 + 4b)\)

\(PQ^2 = (x_1 - x_2)^2 + (y_1- y_2)^2 = 10(a^2 - 6a - 7 + 4b)\)

\(PQ = \sqrt{10(9 + a^2 - 6a - 15 + 4b)} = 4\sqrt{10}\)

\(a^2 - 6a - 7 + 4b = 16\)
\((a - 3)^2 = 32 - 4b >=0\)
b <= 8

IMO D

=>


Assume the intersections are \(P(p, 3p+b)\) and \(Q(q, 3q+b).\)

Then the length of \(PQ\) is

\(\sqrt{(p-q)^2+(3p+b-3q-b)^2}\)

\(= \sqrt{(p-q)^2+(3p-3q)^2}\)

\(= \sqrt{(p-q)^2+9(p-q)^2}\)

\(= \sqrt{10(p-q)^2}\)

\(= \sqrt{10[(p+q)^2-4pq]}\)

\(= 4\sqrt{10} = \sqrt{160} \)

Then we have \(10[(p+q)^2-4pq]=160\) or \((p + q)^2 - 4pq = 16.\)

Since \(p\) and \(q\) are roots of the equation \(x^2 + ax + 4 = 3x + b\) or \(x^2 + (a - 3)x + 4 - b = 0,\) we have \((x - p)(x - q) = x^2 - (p + q)x + pq = x^2 + (a - 3)x + 4 - b, -(p + q) = a – 3\), or \(p + q = -a + 3\) and \(pq = 4 - b.\)

Then \((-a + 3)^2 - 4(4 - b) = 16, (-a + 3)^2 - 16 + 4b = 16\), or \(4b = -(-a + 3)^2 + 32.\)

We have \(b = -(\frac{1}{4})(-a + 3)^2 + 8\) and the maximum value of \(b = 8\) at \(a = 3.\)

Therefore, D is the answer.
Answer: D

Is this question likely to be tested on GMAT? Can someone suggest an easier way to solve it?
Bunuel pls help.

@kinkshook can you please elaborate your method. specially x1+x2 and x1*x2