Answer to Question #223987 in Analytic Geometry for Bless

Question #223987

 Show that the line x + y − 2 = 0 is a tangent to the parabola x2+ 8y = 0 and find the position vector of the point of contact.


1
Expert's answer
2021-08-09T14:51:08-0400
x2+8y=0x^2+8y=0

y=18x2y=-\dfrac{1}{8}x^2

y=14xy'=-\dfrac{1}{4}x

yy1=y(xx1)y-y_1=y'(x-x_1)

y=14x1(xx1)18x12y=-\dfrac{1}{4}x_1(x-x_1)-\dfrac{1}{8}x_1^2

y=18x1x+14x12y=-\dfrac{1}{8}x_1x+\dfrac{1}{4}x_1^2

x+y2=0=>y=x+2x + y − 2 = 0=>y=-x+2

14x1=118x12=2\begin{matrix} -\dfrac{1}{4}x_1=-1 \\ \\ \dfrac{1}{8}x_1^2=2 \end{matrix}

The system has the solution x1=4.x_1=4.


y1=18(4)2=2y_1=-\dfrac{1}{8}(4)^2=-2

The line x+y2=0x+y-2=0 is a tangent to the parabola x2+8y=0x^2+8y=0 at the point (4,2).(4, -2).

The position vector of the point of contact is 4,2.\langle4, -2\rangle.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment