[1] $x^2 + xy + y^2 = 1$
differentiating implicitly by $\frac{d}{dx}$
$2x + x\frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0$
Which simplifies to
[2] $\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$
Now consider a tangent linear equation which passes through a point on the curve and through $(1,1)$
It has the form
[3] $y = mx + c$
To find the points that these equations [1] and [3] intersect we must solve them simultaneously.
Substituting $y$ into [1]
$x^2 + x(mx+c) + (mx+c)^2 = 1$
This simplifies to
$(m^2+m+1)x^2 + (2mc+c)x + c^2 = 1$
using the quadratic formula
[4] $x = \frac{-2mc-c \pm \sqrt{(2mc+c)^2 - 4(m^2+m+1)c^2}}{2(m^2+m+1)}$
Note that there are 2 solutions because there are two points on the curve where this will work.
Also, since we know that equation [3] passes through the point $(1,1)$
we know
$1= m1+c$
So [5] $m = 1 - c$
Se expression [4] becomes
[6] $x = \frac{-2(1 - c)c-c \pm \sqrt{(2(1 - c)c+c)^2 - 4((1 - c)^2+(1 - c)+1)c^2}}{2((1 - c)^2+(1 - c)+1)}$
Lets call these $x_1$ and $x_2$ They have a corresponding $y_1$ and $y_2$
We also know that the gradient of [3] is $m$. This will be equal to the derivative given by [2] at $(x_1, y_1)$ and $(x_2,y_2)$
ie:
[7] $m = \frac{-2x_1 - y_1}{x_1 + 2y_1} = 1 - c$
and
[8] $m = \frac{-2x_2 - y_2}{x_2 + 2y_2} = 1 - c$
Substitute [6] into [7] to get an expression for $y_1$ in terms of c
Then you can substitute $x_1$ and $y_1$ into [1] to find what $c$ is.
From here you can calculate $m$ which will give you the tangent line you are looking for.
A similar process can be used for $x_2$ and $y_2$ there are two tangent lines which will solve this problem.
I'll leave the rest for you
To answer this problem, you must derive a system of linear equations to solve for the three unknowns in your parabola $f(x),$ where $$f(x)=ax^2+bx+c. $$ We know $f(2)=1=4a+2b+c$, $f(4)=3=16a+4b+c$, and $f'(\text{B})=-\frac{1}{4}=4a+b$. Thus, the system $$\begin{cases}4a+2b+c=1\\ 16a+4b+c=3\\ 4a+b=-\frac{1}{4} \end{cases} $$ is established. I will leave it up to you to solve for $f(x).$
Best Answer
HINT: