Find out the answer to "Find equations of both lines through the point..."

Falak Kinney

Answered question

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

Skilled2021-05-18Added 94 answers

The answer to your question:

Jeffrey Jordon

Expert2021-09-09Added 2605 answers

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

$\frac{d y}{d x} = \frac{d}{d x} (x^{2} + x) = \frac{d}{d x} (x^{2}) + \frac{d}{d x} (x)$

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

$\frac{d y}{d x} = 2 \cdot x^{2 - 1} + x^{1 - 1}$

$= 2 x + 1$

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

The slope of the line at $(a, a^{2} + a)$ is $m = \frac{d y}{d x} = 2 a + 1$

Use the slope-intercept form $y = m x + c \Rightarrow y = (2 a + 1) x + C$

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

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

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

$c = - a^{2}$

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

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

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

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

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

$a = 1$ or $a = 5$

From equation (1):

if $a = 1$ , then

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

$\Rightarrow y = (- 2 + 1) x - (- 1)^{2}$

$\Rightarrow y_{1} = - x - 1$

if $a = 5$ , then

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

$\Rightarrow y = (10 + 1) x - 5^{2}$

$\Rightarrow y_{2} = 11 x - 25$

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?