Two orthogonal tangent lines to Ellipse

kunny

In the $xy$ coordinate plane given an ellipse with center $P(p,\ q)$ and has the major axis of the length 4, minor axis of the length 2, these axis are parallel to the $x$ axis, the $y$ axis respectively. Find the range of $p,\ q$ for which two orthogonal tangent lines can be drawn to the ellipse.

**Posted:** Wed Nov 04, 2009 5:15 pm

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Today's calculation of Integral Digest

J.Y.Choi

On the circle passing by four points $(p+2,q+1),(p+2,q-1),(p-2,q+1),(p-2,q-1)$ , you can draw two orthogonal tangent lines to the given ellipse. I can't understand what your problem means.

**Posted:** Fri Nov 06, 2009 7:30 am

kunny

Sorry, I have just edited it.

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Today's calculation of Integral Digest

kunny

**Posted:** Fri Nov 06, 2009 8:58 am

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Today's calculation of Integral Digest

J.Y.Choi

Then, the point $P(p,q)$ must be lying on the circle $x^2+y^2=5$ . Smile

**Posted:** Fri Nov 06, 2009 9:06 am

kunny

How did you solve it?

**Posted:** Fri Nov 06, 2009 9:13 am

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Today's calculation of Integral Digest

igiul

The figure formed by two tangents separated by $pi/2$ and their corresponding normals is a rectangle with sides of 2,1. The tangent lines intersect at the opposite corner from the where the normals intersect @ $P(p,q)$ . Therefore the origin must be exactly $\sqrt{a^2+b^2}=\sqrt{5}$ units from $P(p,q)$ . Essentially, the solution is a locus of points $\sqrt{5}$ units from the origin:

**Posted:** Fri Nov 06, 2009 1:57 pm

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my blawg
$A_f=\pi r^2(2r^2+\alpha^2+\beta^2)$

kunny

That's correct. That's ''director circle''. I would like see how you solved it, or your paper.

By Discriminant or something another solution?

**Posted:** Fri Nov 06, 2009 6:09 pm

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Today's calculation of Integral Digest