Sketch the graph given by the equation $\displaystyle 16x^2-64x+64+a^2y^2-16a^2=0$
I'm not sure how to present the equation to a form suitable for graph sketching(eg. hyperbola, ellipse, circle)
yes, the value of a should be given.Sketch the graph given by the equation $\displaystyle 16x^2+a^2y^2-64x+64-16a^2=0$
if $\displaystyle a\in R$
then two figure are possible.
i)circle ($\displaystyle a^2=4$) and
ii) ellipse ($\displaystyle a^2 != 4$).
check it.
I don't see how that has anything to do with the question you first posted! However, here is a good way to find asymptotes for a hyperbola (which that clearly is). If x and y are both very, very large numbers then so are $\displaystyle \frac{(x-3)^2}{4}$ and $\displaystyle y^2$ so that, in comparison, we can ignore "1". That is, for very, very large x and y, the graph is almost identical with the two straight lines given by $\displaystyle \frac{(x-3)^2}{4}- y^2= 0$.
Now, how about telling us what the problem really is?
The real problem is Sketch the graph given by the equation $\displaystyle 16x^2-64x+64+a^2y^2-16a^2=0$
and the difficulty i am facing is expressing the equation in a form suitable for sketching like a hyperbola, ellipse or circle equation
and i am also allowed to state the intercepts in terms of a too