ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: 95atar on May 25, 2013, 09:20:15 pm

Title: query
Post by: 95atar on May 25, 2013, 09:20:15 pm: find the coordinates of the points on the curve y=4x^2+5 at which the tangents drawn to the curve pass through the point (0,4)
Title: Re: query
Post by: BubbleWrapMan on May 25, 2013, 09:29:28 pm: Generally, if we have a tangent at $(p,f(p))$ , the equation of the tangent is given by

$y=f'(p)(x-p)+f(p) \iff y=f'(p)x+f(p)-pf'(p)$

so the $y$ -intercept of the tangent is $f(p)-pf'(p)$ . If we want this to equal 4, we solve $f(p)-pf'(p)=4\iff 4p^2+5-p(8p)=4\iff 4p^2=1\iff p=\pm\frac{1}{2}$ .

So the required points are $\left(-\frac{1}{2},6\right)$ and $\left(\frac{1}{2},6\right)$ .