Curve $y =4-x^2$ , and line $y$ tangent the curve at $(a,4-a^2)$

Question

I find the equation of the line is $y = -2ax + 4+a^2$. But when I tried to find $a$, the value is $a^2=-4$. How this line can tangent to the curve at point $a$ that is imaginary?

Answer 1

The slope of the tangent line at $(x,4-x^2)$ is $-2x$.

Let $b$ be the $y$-intercept of the tangent line at $(a,4-a^2)$.

Then $-2a(a)+b=4-a^2$, so $b=4+a^2$.

Therefore, the tangent line is $y=-2ax+4+a^2$ for any $a$.

The $y$-intercept is $4+a^2$, and it's never $0$.

Answer 2

You have found the equation of tangent line to the curve at the point $(a, 4-a^2)$ to be $$y = -2ax + 4+a^2$$

Your problem is solved at this point and you do not have to do anything else.

If a value of $a$ is given then you can substitute and find the equation of the tangent line at that given point.

For example if they want to find the tangent line at the point $(1,3)$ then you have $$ y = -2x + 5$$