Question

The equation of the tangent to the parabola $y=(x-3{)}^2$ parallel to the chord joining the points $(3,0)$ and $(4,1)$ is:

• A. $2x-2y+6=0$
• B. $2y-2x+6=0$
• C. $4y-4x+11=0$
• D. $4x-4y=13$

Answer 1

Answer: (D)

Slope of tangent $=\frac{1-0}{4-3}=1$
also $\frac{dy}{dx}=2(x-3)$
${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=2(x,-3)=1\Rightarrow x_1-3=\frac{1}{2}$
$x_1=\frac{7}{2}$
$\therefore y_1={\left(\frac{7}{2}-3\right)}^2=\frac{1}{4}$
Equation of tangent is
$y-\frac{1}{4}=1\left(x-\frac{7}{2}\right)$
$4y-1=2(2x-7)$
$4x-4y=13$