Find the equation of tangent parallel to the secant joining the points on the curve y=x^2y=x2 at x=-1x=1 and x=2x=2?

1 Answer
Mar 15, 2017

Equation is 4x-4y-1=04x4y1=0 and corresponding xx-coordinate is 1/212

Explanation:

y=x^2y=x2 is a parabola. When x=-1x=1, y=1y=1 and when x=2x=2, y=4y=4.

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

As the slope of secant is (4-1)/(2-(-1))=3/3=1412(1)=33=1,

we are seeking a tangent with a slope of 11.

Slope of tangent is given by (dy)/(dx)dydx and as y=x^2y=x2, (dy)/(dx)=2xdydx=2x

Hence, for tangent we should have 2x=12x=1 i.e. x=1/2x=12

but at x=1/2x=12. y=(1/2)^2=1/4y=(12)2=14

Hence we are seeking the tangent at (1/2,1/4)(12,14) with slope 11

and hence equation is y-1/4=x-1/2y14=x12 or 4x-4y-1=04x4y1=0

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