Question

At what point is the tangent to the curve y=x^n parallel to the chord joining (0,0) and (k,k^n)

Answer 1

Answer:

slope of the tangent = y2-y1/ x2-x1

dy/dx=k^n-0/k-0

m=k^n-1

slope of the tangent=slope of the curve

n x^n-1=k^n-1

Answer 2

Step-by-step explanation:

Given: A curve $y=x^{n}$  and a chord joining $(0,0)$ and $(k,k^{n} )$

Let the required point be $(x_{1} , y_{1} )$

Since, the point lies on the curve, it satisfies the curve

⇒$y_{1} =x_{1} ^{n}$

So, the point becomes $(x_{1} , x_{1} ^{n} )$

The slope of a curve at any point:

Slope of a curve at any point is found by differentiating the curve at that point.

Slope at $(x_{1} ,x_{1} ^{n})$ $= \frac{dy}{dx}$ at $x=x_{1}$

$\frac{dy}{dx}$ at $(x=x_{1})$

$= nx_{1} ^{n-1}$

Slope of a chord joining two points:

We have chord joining $(0,0)$ and $(k,k^{n} )$

slope, $m= \frac{k^{n} -0}{k-0}$

              $=k^{n-1}$

According to ques;

$nx_{1} ^{n-1}= k^{n-1}$

⇒$x_{1} = \sqrt[n-1]{\frac{k^{n-1} }{n} }$

Hence, the required point is $(\sqrt[n-1]{\frac{k^{n-1} }{n} },(\sqrt[n-1]{\frac{k^{n-1} }{n} })^{n}$