1
$\begingroup$

I need to find the slope of the tangent of $f(x)=\sqrt{2x-10}$ that pass through point $(0,0)$.

The derivative of $f(x)$ is $1/(\sqrt{2x-10})$.

  • 0
    Now set up the slope intercept form with f'(a) as slope and 0 as y-intercept. – 2017-01-04
  • 0
    Pay attention to the domain! – 2017-01-04
  • 1
    @ArnaldoNascimento it never says that $(0,0)$ should lie on the curve. It is an external point which lies on the tangent. – 2017-01-04
  • 1
    @MattG88 domain is fine. $(0,0)$ is an external point. – 2017-01-04

4 Answers 4

3

Observe that the given curve is a branch of a parabola. Since the tangent line passes through $(0,0)$. Therefore it is of the form $y=mx$. Now find the intersection of this with the given curve by solving \begin{align*} mx & = \sqrt{2x-10}\\ m^2x^2 & = 2x-10\\ m^2x^2-2x+10 & = 0. \end{align*} For this to be a tangent the line should intersect the curve at coincident points, meaning the roots of the above quadratic must be equal. Thus $$4-40m^2 =0 \implies m=\pm \frac{1}{\sqrt{10}}.$$ Keeping in mind that this is the upper branch of the parabola. $m=\frac{1}{\sqrt{10}}$.

enter image description here

  • 0
    :D Good catch! :P – 2017-01-04
  • 0
    @AnuragA What did you do between m^2x^2-2x+10 = 0 and 4-40m^2 = 0 ? – 2017-01-04
  • 0
    That should be @Anurag A who evaluated m by zero determinant.. – 2017-01-04
  • 0
    @Ismael When you solve a quadratic equation $ax^2+bx+c=0$, to have equal roots, the discriminant $b^2-4ac=0$. – 2017-01-05
1

You can choose a point $P(x_P,y_P)$ where the line $y-y_0=m(x-x_0)$ will be tangent to $f(x)=\sqrt{2x-10}$, so if $(x_0,y_0)\equiv O$, we have:

$$y=mx$$ $$m_P={1\over\sqrt{2x_P-10}}$$

thus $y={1\over\sqrt{2x_P-10}}x$ is a line through the origin tangent to $f(x)$ at $P$. Now we know that this line must be tangent to $f(x)$ at $P$ so it must pass through this point, hence:

$$y_P={1\over\sqrt{2x_P-10}}x_P$$

but $y_P=\sqrt{2x_P-10}$, so: $$\sqrt{2x_P-10}={1\over\sqrt{2x_P-10}}x_P\longrightarrow 2x_P-10=x_P\Longrightarrow x_P=10$$

So we get $P(10,\sqrt{10})$ and $m_P={1\over\sqrt{2x_P-10}}={1\over\sqrt{10}}$.

  • 0
    $y={1\over\sqrt{2x-10}}x$ is NOT a line. You should have a different variable for $x$ in the expression for $m$. Moreover the slope should be $1/\sqrt{10}$. – 2017-01-04
  • 0
    @AnuragA yes sure, I corrected it, thanks;) – 2017-01-04
1

$$ y = f(x) \tag1$$ $$ y^2 = 2x -10 \tag2$$ Differentiate

$$ y y^{\prime} = 1 \tag3$$

Take standard form with shift of 5 along $x$ direction.

$$ x= ft^2 + 5 ;\, y = 2 ft \tag4$$

$ y^{\prime}$ = slope

$$ =\frac {dy}{dx} = \frac {dy/dt}{dx/dt} =\frac {2f}{2ft} = \frac1t \tag5 $$

Equation of tangent

$$ \frac {y-2ft}{ft^2 + 5 } = \frac1t \tag6 $$

Since it passes through origin, constant term should vanish. Cross-multiplying for constant terms,

$$ -2 ft^2 = -(5 + ft^2)\rightarrow t = \pm \sqrt 10 \tag7 $$

$$ Slope =1/t = \pm \frac{1}{ \sqrt 10} \tag8$$

Positive sign for tangent in first quadrant, negative in fourth.

0

The tangent is given by $$mx+ b,$$ with $m$ being the derivative of $f$ in $x=a$ ($x$-value to be evaluated).

  • 0
    This hint is very generic and doesn't help solve the problem. Please try to be more specific. – 2017-01-04