the only information given is ellipse equation x^2/17+y^2/8=1, first I tried to substitute y in ellipse equation with m(x-a)+b, then I tried to make determinant zero, getting 2nd order equation of am^2+bm+c=0, then I tried to get the determinant of this equation higher than zero, noting c/a= m1.m2 and m1.m2=-1, but I don't know the rest.
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I can help but before that, you must detail out your effort. Also please use mathjax to write your math. You have been on the site for a while now so you should know. – Math Lover May 18 '21 at 06:02
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1It is well known that the locus of points where two perpendicular tangent to the ellipse cross each other is a circle, called the director circle of the ellipse, whose radius is $r=\sqrt{a^2+b^2}$. – Intelligenti pauca May 18 '21 at 06:56
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As comment by Inteligenti, the locus of theses points is a circle, In figure a a circle is shown and in b an ellipse as a compressed circle. The radius of locus for circle is $R=\sqrt 2 r$ where r is the radius of circle.For ellipse $R=\sqrt {a^2+b^2}$. So if you want to find in your own way this must be final result.
sirous
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