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

Question

3748 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-15T07:00:03

Equation is $4x-4y-1=0$ and corresponding $x$ -coordinate is $1/2$

$y=x^2$ 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 $(-1,1)$ and point $(2,4)$ .

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

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

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

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

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

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

and hence equation is $y-1/4=x-1/2$ or $4x-4y-1=0$

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

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