# Find the equation of tangent parallel to the secant joining the points on the curve y=x^2 at x=-1 and x=2?

Mar 15, 2017

Equation is $4 x - 4 y - 1 = 0$ and corresponding $x$-coordinate is $\frac{1}{2}$

#### Explanation:

$y = {x}^{2}$ is a parabola. When $x = - 1$, $y = 1$ and when $x = 2$, $y = 4$.

Hence, we are seeking a tangent i.e. parallel to the secant joining points $\left(- 1 , 1\right)$ and point $\left(2 , 4\right)$.

As the slope of secant is $\frac{4 - 1}{2 - \left(- 1\right)} = \frac{3}{3} = 1$,

we are seeking a tangent with a slope of $1$.

Slope of tangent is given by $\frac{\mathrm{dy}}{\mathrm{dx}}$ and as $y = {x}^{2}$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

Hence, for tangent we should have $2 x = 1$ i.e. $x = \frac{1}{2}$

but at $x = \frac{1}{2}$. $y = {\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$

Hence we are seeking the tangent at $\left(\frac{1}{2} , \frac{1}{4}\right)$ with slope $1$

and hence equation is $y - \frac{1}{4} = x - \frac{1}{2}$ or $4 x - 4 y - 1 = 0$

graph{(y-x^2)(4x-4y-1)(y-x-2)=0 [-4.71, 5.29, -0.56, 4.44]}