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

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Answer 1 · 2017-03-15T07:00:03

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

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

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 $\frac{4 - 1}{2 - (- 1)} = \frac{3}{3} = 1$ $(4-1)/(2-(-1))=3/3=1$ ,

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

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

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

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

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

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

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=x2y=x^2 at x=-1x=−1x=-1 and x=2x=2x=2?