# Normals using calculus

1. Jul 1, 2017

### Tanishq Nandan

1. The problem statement, all variables and given/known data
Find the interval for a so that (3-a)x+ay+(a^2-1)=0 is normal to the curve xy=4

2. Relevant equations
If two curves intersect each other orthogonally (at right angles)
m1×m2=-1
Slope of a curve y=f(x) at a point(x1,y1) is equal to value of (dy/dx) at x1,y1
3. The attempt at a solution
I simply applied the two formulas stated above.If the given line is normal to the hyperbola,then that means they must intersect orthogonally,i.e we can use the first formula stated above,
finding m1 and m2 from the second formula stated above.
Now,the slope for the line is coming constant (1 - 3/a)
And the slope for the hyperbola is coming (-y1/x1),where x1 and y1 are the points of intersection of the line and the hyperbola.
Now,I can solve the two equations to find x1 and y1 in terms of a and hence,find m2 in terms of a, and then apply m1×m2=-1,but before doing that I noticed that whatever I do,I'm going to be solving an equation,not an inequality,so it will give me nothing but discrete values of a,whereas the question asks for the interval of a.
Hence,my problem..
Is there any other formula I'm missing??

2. Jul 1, 2017

The best I can tell, you get isolated points $a$ for a solution, and the equation for $a$ doesn't look very simple algebraically.

3. Jul 1, 2017

### Tanishq Nandan

And,it isn't simple algebraically,so are you suggesting a graphical approach?
because I tried that but found out that I was even more helpless there.Here,at least I know what I have to do.

4. Jul 1, 2017

Suggestion: Try the point $a=1$. That makes $2x+y=1$ so that $y=-2x$. That line doesn't even intersect the curve $xy=4$. $\\$ It would appear there is an error somewhere=it does happen on occasion.