Conics

**xwrathbringerx** · October 22nd 2010, 04:21 AM

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...

Please help?!?

**Ackbeet** · October 22nd 2010, 04:50 AM

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

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

Multiplying out would give

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

My thought is that the textbook is wrong.

**xwrathbringerx** · October 22nd 2010, 05:26 AM

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?

**Ackbeet** · October 22nd 2010, 05:34 AM

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

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

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

This is the book's solution.

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