1. ## Conics

Hi

For the parabola y = 0.5 (x^2 +1), why is the equation of the tangent at point P (a,b): xa = y + b -1 (in the textbook)?

I tried differentiating the parabola equation to find the gradient (a) and then using y-y1 = m(x-x1) formula but it doesnt turn out to be that...

2. Hmm. $\displaystyle y'(x)=x.$ Hence, for the point $\displaystyle (a,b),$ the tangent line will have the slope you mentioned, namely, $\displaystyle a.$ The equation of the tangent line, using your (correct) equation would be

$\displaystyle y_{\text{tan}}-b=a(x-a).$

Multiplying out would give

$\displaystyle y_{\text{tan}}=ax-a^{2}+b.$

My thought is that the textbook is wrong.

3. Hmmmm the solution is part of this problem: The locus has equation y = 0.5 (x^2+1) Find the gradients of the tangents to this curve which pass through the origin. Could the equation of the tangent be as the solutions say because it passes through the origin?

4. Wait, I might have something here. The point $\displaystyle (a,b)$ satisfies the equation $\displaystyle b=0.5(a^{2}+1),$ right? It's on the parabola. Let's say we solved this equation for $\displaystyle a^{2}.$ We'd get

$\displaystyle 2b-1=a^{2}.$ Plugging that into the equation we had before, we get

$\displaystyle y_{\text{tan}}=ax-(2b-1)+b=ax-2b+1+b=ax-b+1.$

This is the book's solution.