Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find a point on the curve y = (x - 3)^2,where the tangent is parallel to the chord joining the points (3, 0) and (4, 1).

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

 

Answers (1)

Given: Equation of curve, y = (x - 3)^2

Firstly, we differentiate the above equation with respect to x, we get

\begin{aligned} &\frac{d}{d x} y=\frac{d}{d x}(x-3)^{2}\\ &\begin{array}{l} \text { Chain Rule } \\ f(x)=g(h(x)) \\ f^{\prime}(x)=g^{\prime}(h(x)) h^{\prime}(x) \end{array} \end{aligned}

\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \times(\mathrm{x}-3) \times(1-0) \quad$ [using chain rule $]$ \\$\Rightarrow \frac{d y}{d x}=2 x-6
Given tangent to the curve is parallel to the chord joining the points (3, Q) and (4,1)
i.e \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}_{1}-\mathrm{y}_{2}}{\mathrm{x}_{1}-\mathrm{x}_{2}}

\\\Rightarrow 2 x-6=\frac{0-1}{3-4}$ \\$\Rightarrow 2 x-6=\frac{-1}{-1}$ \\$\Rightarrow 2 x-6=1$ \\$\Rightarrow 2 x=7

\Rightarrow \mathrm{x}=\frac{7}{2}=3.5 \in(3,4)
Put x=\frac{7}{2}  in y=(x-3)^{2}, we have
y=\left(\frac{7}{2}-3\right)^{2}=\left(\frac{7-6}{2}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}
Hence, the tangent to the curve is parallel to chord joining the points (3,0) and (4,1) at \left(\frac{7}{2}, \frac{1}{4}\right)

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question