Falak Kinney

2021-05-17

Find equations of both lines through the point (2, ?3) that are tangent to the parabola $y={x}^{2}+x$.
${y}_{1}$=(smaller slope quation)
${y}_{2}$=(larger slope equation)

oppturf

The answer to your question:


Jeffrey Jordon

Differentiate $y={x}^{2}+x$ with respect to x

$\frac{dy}{dx}=\frac{d}{dx}\left({x}^{2}+x\right)=\frac{d}{dx}\left({x}^{2}\right)+\frac{d}{dx}\left(x\right)$

By using the Powe Rule $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$

$\frac{dy}{dx}=2\cdot {x}^{2-1}+{x}^{1-1}$

$=2x+1$

Since $y={x}^{2}+x$, consider a point on the parabola as $\left(a,{a}^{2}+a\right)$.

The slope of the line at $\left(a,{a}^{2}+a\right)$ is $m=\frac{dy}{dx}=2a+1$

Use the slope-intercept form $y=mx+c⇒y=\left(2a+1\right)x+C$

If this line is a tangent to the give parabola, then it passes through $\left(a,{a}^{2}+a\right)$, so

${a}^{2}+a=\left(2a+1\right)a+c$

${a}^{2}+a=2{a}^{2}+a+c$

$c=-{a}^{2}$

Therefore the equation of the tangent line is $y=\left(2a+1\right)x-{a}^{2}$

Since the required tangent passes through the point $\left(2,-3\right)$

$y=\left(2a+1\right)x-{a}^{2}$

$-3=\left(2a+1\right)2-{a}^{2}$

${a}^{2}-4a-5=0$

$a=1$ or $a=5$

From equation (1):

if  $a=1$, then

$y=\left(2a+1\right)x-{a}^{2}$

$⇒y=\left(-2+1\right)x-\left(-1{\right)}^{2}$

$⇒{y}_{1}=-x-1$

if $a=5$, then

$y=\left(2a+1\right)x-{a}^{2}$

$⇒y=\left(10+1\right)x-{5}^{2}$

$⇒{y}_{2}=11x-25$

Do you have a similar question?