The line that goes through the point (0, 3/2 ) and is orthogonal to a tangent line to the part of parabola y =x^2 with x > 0 is y = ax + 3/2.

Question

Find the slope $a$ which is in $y=ax+3/2$

Find the $x$-coordinate of the intersection of the two lines.

I found that a= -1/2 and x coordinate (1,1). I have solved for an hour. I need easy way.step by step — London Jack, Jun 26 '19 at 14:21
$y=ax+\frac{3}{2}$ is the given tangent line to $y=x^2$ for $x>0$? — Dr. Sonnhard Graubner, Jun 26 '19 at 14:21
For someone to give you a method that you’ll understand, you need to tell us more about what it is that you know how to use: Can you use calculus to find tangents and normals? Do you know about pole-polar relationships? Are you familiar with homogeneous coordinates? I could, for instance write up a method that determines $x$ by finding when a certain matrix determinant vanishes, but if you’re not familiar with the methods that lead to it, it’s not going to be of much use to you. — amd, Jun 26 '19 at 20:44
My level is pre-calculus. Firtsly ,I drew a graph and found point of intersection of normal line and tangent line on parabola. Then found slope of tangent line which is 2. Lastly,found slope of normal line -1/2. Coefficient " a " in y=ax+ 3/2 is slope. Answer (1,1). m=a=-1/2 — London Jack, Jun 27 '19 at 08:08

score 2 · Answer 1 · answered Jun 26 '19 at 14:23

2

Hint: Solve the equation $$x^2-ax-\frac{3}{2}=0$$ and set the discriminant equal to zero and determine the variable $a$.

answered Jun 26 '19 at 14:23

score 1 · Accepted Answer · answered Jun 26 '19 at 18:53

Step by step approach (as requested in comment to question):

Call the $x$-coordinate of the intersection of the tangent and normal lines $x_0$.

The $y$-coordinate is then $y_0=x_0^2$.

Call $m$ the slope of the tangent line to $y=x^2$ at $x_0$. What is $m$ in terms of $x_0$?

What is the slope of the normal line at $x_0$? [Hint: it is perpendicular to the tangent line.]

What is the $y$-intercept of the normal line at $x_0$? [Hint: you just found its slope, and $(x_0,x_0^2)$ is a point on it.]

Given that the $y$-intercept of the normal line is $\dfrac32$ and $x_0>0$, what is $x_0$?

Above you got the slope of the normal line at $x_0$ in terms of $x_0$; this is $a$.

Here is how I picture it, with the parabola in red, the tangent line in blue, and the normal line in green:

As I understand from your steps equation will be ➡ 2x=(x^2-3/2)/x-0 Then answer will be uncorrect! — London Jack, Jun 27 '19 at 06:24
equation for what? that's not what I got; please report your answers to each step — J. W. Tanner, Jun 27 '19 at 10:41

2 Answers2