# Find the equations of both tangent lines to the curve defined by y = (1/x^2), x cannot equal 0...

## Question:

Find the equations of both tangent lines to the curve defined by {eq}y = \frac1{x^2} {/eq}, x cannot equal 0 from the external point P(0,1).

## Tangent Line:

The slope of a tangent line to the graph of {eq}y = f\left ( x \right ) {/eq} at {eq}x=a {/eq} is {eq}f{}'\left ( a \right ) {/eq}.

The equation of the tangent line at {eq}\left ( a,f\left ( a \right ) \right ) {/eq} is {eq}y-f\left ( a \right ) =f{}'\left ( a \right )\left ( x-a \right ) {/eq}.

## Answer and Explanation:

Let {eq}\left ( x_{1},y_{1} \right ) {/eq} be the point of contact of the tangent drawn form the point {eq}P\left ( 0,1 \right ) {/eq} and the curve {eq}y=\frac{1}{x^2} {/eq}. Then

{eq}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{x^3} {/eq}.

Therefore, slope of the tangent at {eq}\left ( x_{1},y_{1} \right ) {/eq} is {eq}\frac{-2}{x_{1}^3} {/eq}.

Since {eq}\left ( x_{1},y_{1} \right ) {/eq} is a point on the curve {eq}y=\frac{1}{x^2} {/eq}, we have {eq}y_{1}=\frac{1}{x_{1}^2} {/eq}.

Therefore, slope of the tangent which joins the points {eq}\left ( 0,1 \right ) {/eq} and {eq}\left ( x_{1},\frac{1}{x_{1}^2} \right ) {/eq} is

{eq}\frac{\frac{1}{x_{1}^2}-1}{x_{1}-0} = \frac{1-x_{1}^2}{x_{1}^3} {/eq}.

{eq}\begin{align*} \therefore \frac{1-x_{1}^2}{x_{1}^3} &= \frac{-2}{x_{1}^3} \\x_{1}^2-1 &= 2 \\ \Rightarrow x_{1}^2 &= 3 \\ \Rightarrow x_{1} &= \pm \sqrt{3} \end{align*} {/eq}

Therefore, the equations of the required tangents are {eq}y-1 = \pm \sqrt{3}x {/eq}.

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from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11