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

##### 1 Answer

Mar 15, 2017

Equation is

#### Explanation:

Hence, we are seeking a tangent i.e. parallel to the secant joining points

As the slope of secant is

we are seeking a tangent with a slope of

Slope of tangent is given by

Hence, for tangent we should have

but at

Hence we are seeking the tangent at

and hence equation is

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