Parabolas: What's the Point?

If all the members of the family of curves $f_{\alpha}(x)=2x^2+\alpha x +3 \alpha$ pass through the same point $(a,b)$ , then $f_{\alpha}(a) = f_{\beta}(a) = b$ , for all real $\alpha$ and $\beta$ , such that $\alpha \neq \beta$ .

$2a^2+\alpha a + 3\alpha = 2a^2+\beta a + 3\beta$

$(\alpha - \beta)a = -3(\alpha -\beta)$

$a=-3$

$f_{\alpha}(-3) = 2(-3)^2-3\alpha + 3\alpha=18$

Thus, $(a,b)=(-3,18)$ and $a+b=15$ .

$\text{Since the value of }f_\alpha(a) \text{ is independent of } \alpha \text{, the second and} \\ \text{third terms in } f_\alpha(x) \text{ must cancel when the function is} \\ \text {evaluated at } a. \text{ Thus } a=-3 \text {, then } b=f_\alpha(-3)=18, \\ \text{so } a+b=15.$

We know that for all $\alpha$ , $2a^2 + \alpha a + 3\alpha = b$ .

Let the polynomial $p(\alpha) = (a+3)\alpha$ . Rearranging the above equation gives $p(\alpha) = b - 2a^2$ , a constant. Therefore, $\deg[p(\alpha)] = 0$ , so $a$ must be $-3$ . Furthermore, $p(\alpha) = (-3+3)\alpha = 0 = b-2a^2 = b-2(-3)^2 = b-18$ , so $b = 18$ . The answer to our problem is therefore $-3 + 18 = \boxed{15}$

2a^2 + αa + 3α = b

Therefore, there cannot be an α term in the equation.

α(a+ 3) = 0

a = -3

b = 2(-3)^2 = 18

a + b = 15

The parabola cuts (a,b) for any real number α so α would not change the value of b.

So f_α(x)=2 x^2+α x+3α can be written as b = 2 a^2 + α a + 3α and as α would not affect value of b then a + 3 = 0 i.e. a = -3 and b = 2 a^2 = 2 (-3)^2 = 18 .Hence a + b = 15

fa(x)=fb(x), where a not equal b. 2x^2+ax+3a=2x^2+bx+3b solving the equation, (a-b)x+3(a-b)=0. therefore x=-3. then fa(x)=18. x+fa(x)=-3+18=15

We want fα(x) to be constant in terms of α, so x=-3 gives -3+18=15

take any $\beta \neq \alpha$ . Given that both and $f_{\alpha}(x)$ and $f_{\beta}(x)$ passes through $(a,b)$ . Hence, $f_{\alpha}(a) = f_{\beta}(a)$ Which leads to $(\alpha-\beta)(a+3)=0$ or $a=-3$ , since $\beta \neq \alpha$ . then $b = f_{\alpha}(-3) = 18$ So, $a+b=15$

We have, 2x^2 + ɑx + 3ɑ= 2x^2 + (x+3)ɑ

Now for this value to be independent of ɑ, we must have x+3= 0, or x= -3.

If x= -3, then 2x^2 + ɑx + 3ɑ= 18.

Both these values are constant and independent of ɑ.

So for any value of ɑ, this line must pass through (-3, 18)

Thus, a= -3, b= 18 and a+b= -3+18= 15.

We want to find $a$ such that $f_\alpha(a) = b$ for some constants $(a,b)$ that do not depend on $\alpha$ . Taking the derivative in respect to $\alpha$ at both sides, we find that $\frac{\partial}{\partial \alpha} f_\alpha(a) = 0.$ Then, we find that $a+3=0$ , which is equivalent to $a = -3$ . Finally, we can check that $f_\alpha (-3) = 18$ , for any real number $\alpha$ .

Method 1 : we can approach this question by observing that :

$2{ x }^{ 2 }\quad -\quad \alpha (x+3)=0$ .

Represent Family of Parabola ( $y=2{ x }^{ 2 }$ .) and line ( $y=x+3$ .)

Which always Pass through an Fixed Point that is intersection of Parabola and the Line which on calculation comes out to be as $(-3,18)$ .

Method - 2

Mathematical Interpretation That we made after reading the question is That :

" For a unique Point $P(a,b)$

${ f }_{ \alpha }(a)\quad =\quad b\quad \quad \forall \quad \quad \alpha \quad \in \quad R$ . "

Now This equation Becomes:

$(2{ a }^{ 2 }-b)\quad +\quad \alpha (a+3)=0$ .

Now Let $g(\alpha )$ . be a polynomial of degree '1' .

such That $g(\alpha )=(2{ a }^{ 2 }-b)\quad +\quad \alpha (a+3)$ .

Since This Polynomial has degree '1' so it has at-most '1' real root. But according to question $g(\alpha )=0$ . has infinite Roots because it is true for every $\alpha \quad \in \quad R$ .

Hence It must be an identity so it must be independent of $\quad \alpha$ .

Hence it's Coefficient's must be $0$ .

so we get an unique Pair of (a,b)

i.e $a=-3\quad \& \quad b=18$ .

Q.E.D

when α=3, fα(x)=2x2+3x+9 when x =a, b=2a2+3a+9 ( 1st equation)

when α=4 fα(x)=2x2+4x+12 when x=a b=2a2+4a+12 (2nd equation)

then solve it simultaneously to get your answer

As α is real, let us take just α = 1, 2, 3 .... Now write down fα(x) values for α = 1 and 2. Given that fα(x) passes through the point (a,b), which means fα(a) = fα(b) for α = 1 and 2 On solving we get a+3 = 0 This gives that a = -3 Now let us substitute the value of 'a' in either of the equations fα(x) for α = 1 and 2, we get the value of b as 18. Therefore the point (a, b) = (-3, 18) hence the value of a+b = -3 + 18 = 15. Thats it......!

as it is valid for every value of alpha there fore first we put alpha=1 ant (a,b) then we get b=2a^2 and then put alpha=1 then next equation will be obtained b=2a^2+3a+3 by simultaniously solving these two equation we get a=-3 and b=18 and their sum is 15

Since the parabola passes through the common point $(a,b)$ , then: $b = 2a^2 + αa + 3α$ Now since $α$ is any real number, then let us consider the case when it is $0$ and $1$ case $α=0$ then $b = 2a^2$ , case $α=1$ then $b = 2a^2 + a + 3$ , substituting $b = 2a^2$ we get $a = -3$ , thus $b = 18$ , therefore ( a + b = -3 + 18 = 15

Since the point is independent of $\alpha$ , writing $f_\alpha (x) = 2x^2 + \alpha ( x + 3)$ , we will require $x=-3$ . This gives $f_\alpha (-3) = 18$ , so $f_\alpha$ must pass through the point $(-3, 18)$ . Hence, $a + b = -3 + 18 = 15$ .

For intersection the required condition is $2x^{2}+\alpha x+3\alpha = 2x^{2}+(\alpha +n)x+3(\alpha +n)$ For any real number $n$ .By cancellation we get $x=-3$ . By putting that we find that the value of y for which this occurs is independent of $\alpha$ and that value of y is $18$ .

For there to be a point that has the same coordinates across all parabolas, these must be independant of $\alpha$ . In other words, all the terms with alphas must cancel out.

$a \alpha +3 \alpha = 0$

$\alpha (a+3) = 0$

$a + 3 = 0 ( \alpha \neq 0)$

a = -3

$b = 2 \times (-3)^2 -3 \alpha +3 \alpha$

$b = 2 \times 9 + 0$

b = 18

$a + b = -3 + 18 = 15$ as for the case where $\alpha = 0$ :

Since the alpha terms cancel out anyway you still get the same solution.

Take alpha common, we get 2x^2 +a (x+3). If x=-3 then the term with 'a ' becomes 0 and corresponding 'y' is 18. We can also put a=0 once and a=1 then solve the two equations.

Since ( a,b ) is a point of our function we can rewrite it as ( a,f(a) ).

We know that

$f _{ \alpha }\left( a \right) =b\quad \forall \alpha \in R$

and then

$f _ {\alpha} \left( a \right) =b= f _ {\alpha +\varepsilon} \left( a \right) \quad \forall \alpha ,\varepsilon \in R$

Solving that equation we find $a=-3$ and $b=f _{1}(a)=18$ so the solution of the problem is

$\boxed{a+b=15}$

"For any real number $α$ , the parabola $f_\alpha(x) =f_\alpha(x) = 2x^{2} + \alpha x + 3\alpha$ passes through the point $(a,b)$ "

That statement gives us a system consists of infinite equations which can be roughly written as following:

$\alpha = 0 \Rightarrow f_\alpha(x) =f_{0}(x) = 2x^{2} + (0)x + 3(0)$ $\alpha = 1 \Rightarrow f_\alpha(x) =f_{1}(x) = 2x^{2} + (1)x + 3(1)$ $\alpha = 2 \Rightarrow f_\alpha(x) =f_{2}(x) = 2x^{2} + (2)x + 3(2)$ $\text{(...)}$ $\alpha = n \Rightarrow f_\alpha(x) =f_{n}(x) = 2x^{2} + (n)x + 3(n)$ $\alpha = n+1 \Rightarrow f_\alpha(x) =f_{n}(x) = 2x^{2} + (n+1)x + 3(n+1)$ $\alpha = n+2 \Rightarrow f_\alpha(x) =f_{n}(x) = 2x^{2} + (n+2)x + 3(n+2)$ $\text{(and so on...)}$

It is also stated that for every equation in that system, there exists a particular value of $x = a$ which will always be mapped onto particular value of $y = b$ . Both a and b are a pair of fixed values for all equations.

So, It means $b = f_\alpha(a) = 2a^{2} + \alpha a + 3(\alpha)$

which can be expressed as following:

$\alpha = 0 \Rightarrow b = f_\alpha(a) =f_{0}(a) = 2a^{2} + (0)a + 3(0)$ $\alpha = 1 \Rightarrow b = f_\alpha(a) =f_{1}(a) = 2a^{2} + (1)a + 3(1)$ $\alpha = 2 \Rightarrow b = f_\alpha(a) =f_{2}(a) = 2a^{2} + (2)a + 3(2)$ $\text{(...)}$ $\alpha = n \Rightarrow b = f_\alpha(a) =f_{n}(a) = 2a^{2} + (n)a + 3(n)$ $\alpha = n+1 \Rightarrow b = f_\alpha(a) =f_{n}(a) = 2a^{2} + (n+1)a + 3(n+1)$ $\alpha = n+2 \Rightarrow b = f_\alpha(a) =f_{n}(a) = 2a^{2} + (n+2)a + 3(n+2)$ $\text{(and so on...)}$

From that, we may express them into: $b = f_\alpha(a) =f_{0}(a) = f_{1}(a) = f_{2}(a) = ... = f_{n}(a) = f_{(n+1)}(a) = ...$

Given $\psi$ as any real number, the expression can be inductively generalized as

$f_\alpha(x) = f_{\alpha+\psi}(x) \text{ for x = a}$

And all of them can be concluded as:

$\{ \forall \alpha , \psi \in \mathbb{R} , \exists x \in \mathbb{R} | f_\alpha(x) = 2x^{2} + \alpha x + 3 \alpha , x = a \implies f_\alpha(x) = f_{\alpha + \psi}(x) \}$

Now we can do mathematical induction in order to find the value of a.

$f_\alpha(a) = f_{\alpha+\psi}(a)$ $2a^{2} + (\alpha)a + 3(\alpha) = 2a^{2} + (\alpha+\psi)a + 3(\alpha+\psi)$ $0 = 2a^{2} + (\alpha+\psi)a + 3(\alpha+\psi) - (2a^{2} + (\alpha)a + 3(\alpha))$ $0 = (\alpha+\psi)a - (\alpha)a + 3(\alpha+\psi) - 3(\alpha)$ $0 = \alpha a - \alpha a + \psi a + 3\alpha - 3\alpha +3\psi$ $0 = \psi a +3\psi$ $0 = \psi(a +3)$ $0 = a +3$ $a = -3$

by knowing $a = -3$ , we may calculate $b$ using any value of $\alpha$

if we choose $\alpha = 0$ , then:

$b = f_\alpha(a) =f_{0}(a) = f_{0}(-3) = 2(-3)^{2} + (0)(-3) + 3(0)$ $b = f_\alpha(a) =f_{0}(a) = f_{0}(-3) = 2(9) + 0 + 0$ $b = f_\alpha(a) =f_{0}(a) = f_{0}(-3) = 18$ $b = 18$

Now we know the value of $a$ which is $-3$ and the value of $b$ which is $18$ . The value of $a+b$ is $-3+18 = 15$ .