Hello Brandon,
1.) We are given the curve:
$$y=2x^2-kx+3$$ where $k$ is a constant.
Let point $A$ be given by:
$$\left(x_A,y_A \right)=\left(x_A,2x_A^2-kx_A+3 \right)$$
The slope of the line passing through the points $A$ and $B$ is:
$$m=\frac{\left(2x_A^2-kx_A+3 \right)-(1)}{\left(x_A \right)-(5)}=\frac{2x_A^2-kx_A+2}{x_A-5}$$
Also, given that:
$$m=\left.\frac{dy}{dx} \right|_{x=x_A}=4x_A-k$$
We may then state:
$$\frac{2x_A^2-kx_A+2}{x_A-5}=4x_A-k$$
Multiplying through by $$x_A-5$$, we obtain:
$$2x_A^2-kx_A+2=4x_A^2-(20+k)x_A+5k$$
Combining like terms, we may arrange this as:
$$5k=2+20x_A-2x_A^2$$
Hence:
$$k=\frac{2}{5}\left(1+10x_A-x_A^2 \right)$$
2.) We are given the curve:
$$y=x+x^2$$
i) First we find the slope of the tangent line at $x=a$ is:
$$m=\left.\frac{dy}{dx} \right|_{x=a}=2a+1$$
And this tangent line must pass through the point:
$$\left(a,a+a^2 \right)$$
Thus, using the point-slope formula, we find the equation of the tangent line is given by:
$$y-\left(a+a^2 \right)=(2a+1)(x-a)$$
Distributing on the right side, we get:
$$y-\left(a+a^2 \right)=(2a+1)x-2a^2-a$$
Adding $a+a^2$ to both sides, we get the tangent line in slope-intercept form:
$$y=(2a+1)x-a^2$$
ii) If this tangent line passes through the point $(-2,3)$, then we must have:
$$3=(2a+1)(-2)-a^2$$
Now we may solve for $a$. Arranging the quadratic in $a$ in standard form, we obtain:
$$a^2+4a+5=0$$
We see the discriminant is negative, and so we conclude there are no real values of $a$ for which a tangent line to the given quadratic curve will pass through point $B$.
iii) There are no such possible tangent lines as we found in part ii).
I suspect that point $P$ was incorrectly given. I will consider the two following cases:
a) Point $P$ is supposed to be $(-2,-3)$. Then part ii) becomes:
$$-3=(2a+1)(-2)-a^2$$
Now we may solve for $a$. Arranging the quadratic in $a$ in standard form, we obtain:
$$a^2+4a-1=0$$
Application of the quadratic formula yields:
$$a=-2\pm\sqrt{5}$$
And so the two tangent lines would be given by:
$$y=\left(2\left(-2\pm\sqrt{5} \right)+1 \right)x-\left(-2\pm\sqrt{5} \right)^2=\left(-3\pm2\sqrt{5} \right)x-\left(9\pm4\sqrt{5} \right)$$
b) Point $P$ is supposed to be $(2,3)$. Then part ii) becomes:
$$3=(2a+1)(2)-a^2$$
Now we may solve for $a$. Arranging the quadratic in $a$ in standard form, we obtain:
$$a^2-4a+1=0$$
Application of the quadratic formula yields:
$$a=2\pm\sqrt{3}$$
And so the two tangent lines would be given by:
$$y=\left(2\left(2\pm\sqrt{3} \right)+1 \right)x-\left(2\pm\sqrt{3} \right)^2=\left(5\pm2\sqrt{3} \right)x-\left(7\pm4\sqrt{3} \right)$$
