Question

Find a point on the parabola y= (x-3)^2 where the tangent is parallel to the chord joining (3,0) and (4,1)

Answer 1

Answer:

The required point is $(\frac{7}{2},\frac{1}{4})$

Step-by-step explanation:

Formula used:

Slope of the line joining $(x_1,y_1)\:and\:(x_2,y_2)\:is\:\frac{y_2-y_1}{x_2-x_1}$

Let (a,b) be a point on the parabola $y=(x-3)^2$ at which the tangent is parallel to the chord joining (3,0) and (4,1)

Then, $b=(a-3)^2$............(1)

slope of the chord joining(3,0) and (4,1)

$=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{1-0}{4-3}$

$=\frac{1}{1}$

$=1$

$y=(x-3)^2$

Differentiate with respect to x

$\frac{dy}{dx}=2(x-3)$

slope of tangent

$=\frac{dy}{dx}_(a,b)=2(a-3)$

As per given tangent at (a,b) is parallel to

the chord joining (3,0) and (4,1)

Therefore, their slopes are equal

$2(a-3)=1$

$a-3=\frac{1}{2}$

$a=\frac{1}{2}+3$

$a=\frac{1+6}{2}$

$a=\frac{7}{2}$

put $a=\frac{7}{2}$ in (1) we get,

$b=(\frac{7}{2}-3)^2$

$b=(\frac{7-6}{2})^2$

$b=(\frac{1}{2})^2$

$b=\frac{1}{4}$

The required point is $(\frac{7}{2},\frac{1}{4})$