I have an ellipse given by:
$$9x^2+16y^2=144$$
and I need to determine the point on the ellipse that is closest to the line passing through the point $A=(5,2)$ and is perpendicular to the line $$l: 𝑦 = 7 + x.$$
I try to set the ellipse equal to the tangent line first:
$$9x^2+16y^2-144=7 + x$$
But this cannot be solved.
Any ideas appreciated..
Best Answer
Hint:
Write a perpendicular to $y=x+7$ which passes through $A$
So you need to find a line parallel to $\ell$ which is tangent to ellipse.
So you have to solve an equation $$9x^2 +16(-x+n)^2 = 144$$ which has only one solution in $x$. This means that it discriminant is $0$. So you get an $n$ and then $x$ of a touching point $D$. Then calculate $d(D,\ell)$ and you are done.