One more q on conic

[Sonal7]

I hope I havent bombared with a list. Its one part of a question that I am not able to do. I cant do part (d).

 

[Sonal7]

I have included the answer to the previous part. Its a nice question.
I am now trying to get the equation y^2=4ax. First I get a value for t in terms of Y

t=Y/2a

sub in at^2=2(X-a). So I get Y^2/4a. But why would 16Y^2 =4aX is also parabola ? It looks like it on desmos.

 

[Sonal7]

 

[Deleted member 4993]

I have included the answer to the previous part. Its a nice question.
I am now trying to get the equation y^2=4ax. First I get a value for t in terms of Y

t=Y/2a

sub in at^2=2(X-a). So I get Y^2/4a. But why would 16Y^2 =4aX is also parabola ? It looks like it on desmos.

Because it has properties of a parabola - albeit a horizontal parabola (where as y = 4*a*x² becomes a vertical parabola).

 

[Sonal7]

Because it has properties of a parabola - albeit a horizontal parabola (where as y = 4*a*x² becomes a vertical parabola).

I am not convinced its a parabola. Parabola points are equidistant from a focus and directrix and i cant see it on desmos. I am sure I am missing what you are saying. so you are telling me that you can stretch a parabola and its still a prabola.

 

[Sonal7]

Because it has properties of a parabola - albeit a horizontal parabola (where as y = 4*a*x² becomes a vertical parabola).

yes its a parabola, even though the a will be different if you have to have a 4 before the ax^2