Question

class 12 application of derivative to find the tangent point on a curve​

Answer 1

The straight line is $y=k.$

It's slope is,

$\longrightarrow m=\dfrac{dy}{dx}$

$\longrightarrow m=\dfrac{d}{dx}(k)$

Since $k$ is a constant,

$\longrightarrow m=0\quad\quad\dots(1)$

The curve is $y=x^2-3x+2.$

Slope of tangent to this curve is,

$\longrightarrow m=\dfrac{dy}{dx}$

$\longrightarrow m=\dfrac{d}{dx}\big(x^2-3x+2\big)$

$\longrightarrow m=2x-3\quad\quad\dots(2)$

Equating (1) and (2),

$\longrightarrow 2x-3=0$

$\longrightarrow x=\dfrac{3}{2}$

And,

$\longrightarrow y=x^2-3x+2$

$\longrightarrow y=(x-1)(x-2)$

$\longrightarrow y=\left(\dfrac{3}{2}-1\right)\left(\dfrac{3}{2}-2\right)$

$\longrightarrow y=-\dfrac{1}{4}$

Hence the coordinates of the point is,

$\longrightarrow\underline{\underline{(x,\ y)=\left(\dfrac{3}{2},\ -\dfrac{1}{4}\right)}}$