<s>[INST] A not so hard basic calculus problem? But it appears to be very lengthy Find the coordinates of the two points on the curve $y=4-x^2$ whose tangents pass through the point $(-1,7)$.
My work:
Let the two points be $(a,b)$ and $(c,d)$. And $\frac{dy}{dx}=-2x$, so the gradients of the tangents are $-2a$ and $-2c$. We get $\frac{7-d}{-1-c}=-2c\implies 7-d=2c+2c^2$ and $\frac{7-b}{-1-a}=-2a\implies 7-b=2a+2a^2$. Now we have 2 equations and 4 unknowns. So we need two more equations.
We have the equations of the tangents are $y=-2cx+2c^2+d$ and $y=-2ax+2a^2+b$. The solutions of these two equations are $(-1,7)$, putting $(-1,7)$ in we get $7(a-c)=-2ac(a-c)+ad-bc$ and $2(a-c)=-2(a-c)(a+c)+d-b$.
We can also equate $4-x^2=-2cx+2c^2+d$ and $4-x^2=-2ax+2a^2+b$, then we get $a^2-c^2=d-b$. From here, we still have to substitute and do some algebra to get $(a,b)$ and $(c,d)$.
Is my working correct? But I feel like the working is too lengthy. Is there any faster way?
Many thanks!
 [/INST] Your work is correct. Maybe you can shorten it by assuming that the points of tangency is $P=(t,4-t^2)$. Then slope at $P$ is $m_P=-2t$. Slope of line $LP=\dfrac{7-(4-t^2)}{-1-t}=\dfrac{3+t^2}{-(1+t)}$. Now this must be equal to $m_P$.
$$\dfrac{3+t^2}{-(1+t)}=-2t.$$
Solving this gives
$$t^2+2t-3=0.$$
Now solve for the two values of $t$. 
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