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The question:

y=3.48x^2, determine the positionson the curve (x,y) of the left most tangent and the right most tangent that pass through the point (-5.67, -44.75).

If there is no tangent for either the left or right, just say none.

Round to 4 dec. places.

Thanks

math
Oli96  Aug 22, 2014

Best Answer 

 #3
avatar+18712 
+5

 

Find the most left and right tangent

$$\boxed{y=a*x^2\quad a=3.48}\\
\mbox{Point }(x_p,y_p) \text{ : } \left( x_p = -5.67 \mbox{ , }y_p =-44.75 \right)\\
\text{Find the tanget Point } (x_t, y_t) \\ \\
y_t = a*x_t^2\\
\text{Slop of } y=a*x^2 \text{ is } y'= 2a*x_t \\
\text{Slop of the line through Point p is } y' = \dfrac{y_t-y_p}{x_t-x_p}\\
\text{the slops must be equal: } y' = 2a*x_t = \dfrac{y_t-y_p}{x_t-x_p}\\\\
\Rightarrow 2a*x_t*(x_t-x_p)=y_t-y_p \quad | \quad y_t=a*x_t^2 \\
2a*x_t*(x_t-x_p)=a*x_t^2-y_p \\
2a*x_t^2 - 2a*x_t*x_p=a*x_t^2-y_p \\
2a*x_t^2 -a*x_t^2 - 2a*x_t*x_p + y_p = 0 \\
\boxed{a*x_t^2 - 2a*x_t*x_p + y_p = 0 }\\$$

$$\Rightarrow \boxed{ x_{t_{1,2}}=x_p\pm\sqrt{x_p^2-\dfrac{y_p}{a}}
\qquad y_{t_{1,2}}=a*x_{t_{1,2}}^2}$$

$$x_{t_1}=-5.67+\sqrt{(-5.67)^2-\left(\dfrac{-44.75}{3.48}\right) }\\
x_{t_1}=-5.67+6.70880730103 = 1.03880730103 \\
y_{t_1}=3.48*1.03880730103 ^2 = 3.75533971815$$
right most tangent

$$x_{t_2}=-5.67-\sqrt{(-5.67)^2-\left(\dfrac{-44.75}{3.48}\right) }\\
x_{t_2}=-5.67-6.70880730103 = -12.3788073010\\
y_{t_2}=3.48*(-12.3788073010)^2 = 533.257348282$$
 left most tangent 

heureka  Aug 22, 2014
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2+0 Answers

 #2
avatar+78618 
+5

This one is a little tough!!

The slope of a tangent line to the given parabola at any point is just y' = 6.96x

Now, what we're looking for is at least one point on the parabola where the line through  (-5.67, - 44.75) is tangent to that point (or points).

Let's call the point(s) on the parabola (x, 3.48x^2). And the slope of the tangent line at that point is just 6.96x.

So, using this point on the parabola and the point (-5.67 , - 44.75), we have that, using the slope "formula,"

(3.48x^2 + 44.75) / (x + 5.67) = 6.96x

(3.48x^2 + 44.75)  /(x + 5.67) - 6.96x = 0

And solving this equation using the onsite calculator, we have....

$${\frac{\left({\mathtt{3.48}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{44.75}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5.67}}\right)}}{\mathtt{\,-\,}}{\mathtt{6.96}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3\,406\,662\,741}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{49\,329}}\right)}{{\mathtt{8\,700}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{3\,406\,662\,741}}}}{\mathtt{\,-\,}}{\mathtt{49\,329}}\right)}{{\mathtt{8\,700}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{12.378\: \!807\: \!301\: \!025\: \!932}}\\
{\mathtt{x}} = {\mathtt{1.038\: \!807\: \!301\: \!025\: \!932}}\\
\end{array} \right\}$$

And the equation of the line that goes through (-5.67, -44.75) and touches the parabola at 1.038807301025932 is given by

 

y +44.75 = 6.96(1.038807301025932)(x + 5.67)

y = 7.23009881514048672x + 40.9946602818465597024 - 44.75

y = 7.23009881514048672x -3.7553397181534402976 ........ and rounding, we have

y = 7.23x - 3.755

 

 

And the equation of the line that touches the graph at - 12.378807301025932 is given by

y + 44.75 = 6.96( - 12.378807301025932)(x + 5.67)

y = -86.15649881514048672x -488.5073482818465597024 - 44.75 

y = -86.15649881514048672x - 533.2573482818465597024  .....and rounding, we have

y = -86.156x - 533.257

 

A graph of the solution is found here.........https://www.desmos.com/calculator/zoxdwcc43a

 

Whew!!!   That one was pretty challenging!!!

 

CPhill  Aug 22, 2014
 #3
avatar+18712 
+5
Best Answer

 

Find the most left and right tangent

$$\boxed{y=a*x^2\quad a=3.48}\\
\mbox{Point }(x_p,y_p) \text{ : } \left( x_p = -5.67 \mbox{ , }y_p =-44.75 \right)\\
\text{Find the tanget Point } (x_t, y_t) \\ \\
y_t = a*x_t^2\\
\text{Slop of } y=a*x^2 \text{ is } y'= 2a*x_t \\
\text{Slop of the line through Point p is } y' = \dfrac{y_t-y_p}{x_t-x_p}\\
\text{the slops must be equal: } y' = 2a*x_t = \dfrac{y_t-y_p}{x_t-x_p}\\\\
\Rightarrow 2a*x_t*(x_t-x_p)=y_t-y_p \quad | \quad y_t=a*x_t^2 \\
2a*x_t*(x_t-x_p)=a*x_t^2-y_p \\
2a*x_t^2 - 2a*x_t*x_p=a*x_t^2-y_p \\
2a*x_t^2 -a*x_t^2 - 2a*x_t*x_p + y_p = 0 \\
\boxed{a*x_t^2 - 2a*x_t*x_p + y_p = 0 }\\$$

$$\Rightarrow \boxed{ x_{t_{1,2}}=x_p\pm\sqrt{x_p^2-\dfrac{y_p}{a}}
\qquad y_{t_{1,2}}=a*x_{t_{1,2}}^2}$$

$$x_{t_1}=-5.67+\sqrt{(-5.67)^2-\left(\dfrac{-44.75}{3.48}\right) }\\
x_{t_1}=-5.67+6.70880730103 = 1.03880730103 \\
y_{t_1}=3.48*1.03880730103 ^2 = 3.75533971815$$
right most tangent

$$x_{t_2}=-5.67-\sqrt{(-5.67)^2-\left(\dfrac{-44.75}{3.48}\right) }\\
x_{t_2}=-5.67-6.70880730103 = -12.3788073010\\
y_{t_2}=3.48*(-12.3788073010)^2 = 533.257348282$$
 left most tangent 

heureka  Aug 22, 2014

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