Question

Show that the curves 2x^2+y^2=3 and x=y^2 are orthogonal

Answer 1

Do you know the meaning of orthogonal? It means that at the point of intersection, tangent to one curve is a normal to the other. 2x^2 + y^2=3
or, 2(x^2/3)+y^2/3=1
This is an equation of an ellipse.
x=y^2 is an equation of a parabola.
Can you solve thse two equations and get the point of intersection?
After getting the point of intersection, write the normal equation of one curve and the tangent equation of the other at that point. If they are the same equation, they are orthogonal.
Do u get it?
Equation of tangent:
\[(y-y1)=dy/dx(x-x1)\]
Equation of normal:
\[(y-y1)=-dx/dy(x-x1)\]
where x1,y1 is the point of intersection.
Find dy/dx of each curve, and then use them in the above equations.
If you dont understand, tell me

Answer 2

Substitute x=y^2 in the first equation.
2y^4+y^2-3=0;
Can you solve this quadratic?

Answer 3

Yes, good. That is the value of y^2. But we need y. So we will square root it. If we square root 1, we get \[y=\pm1\]. But if we square root the other, we get an imaginary quantity. So there is no need of considering that.
Now, we know that the y-coordinate of the point of intersection is either +1 or -1. Now, we shall find the x-coordinate. Put y=+-1 in any of the curve, and you will get x=1.
So, y1=1. x1=1.
Now, find dy/dx of each of the curves. Can you do it?

Answer 4

No problem, tell me if you get stuck anywhere