Jose luis M.

asked • 12/18/22

Find the locus of the foot of the perpendicular from Q to the line MN, when Q describes the circle x2+y2=1

Be A and B points of contact of the hyperbola, which equation is canonical one, with the tangents through Q(h,k).

S is a circle through O, A and B, that still intercepts the hyperbola in M and N.

Find the locus of the foot of the perpendicular from Q to the line MN, when Q describes the circle x2+y2=1.

Suggestion: the circle through O,A,B, belongs to the pencil determined by the hyperbola and the lines MN

(polar of Q) and r, an arbitrary line.

Answer: (a2+b2)2 (x2/a2 + y2/b2)=1


Mark M.

Is this one problem? If so are there two circles Q and S? What is the equation of the hyperbola? Revew your post for accuracy.
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12/18/22

Jose luis M.

Hi. Yes, it is just one problem. The equation of the hyperbola is the canonical one: x^2/a^2 - y^2/b^2= 1. There is the first circle that passes through the origin O and the points of contact the tangents with the hyperbola. And the second circle x^2 + y^2 =1 that it is course that the point takes. Sorry for any misunderstanding.
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12/19/22

Mark M.

OK, I am going to try to draw and label a diagram. Did you?
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12/19/22

Mark M.

Too many undefined - A, B, M, N, O, and Q/
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12/19/22

Jose luis M.

yes, it's a literal question.
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12/19/22

Jose luis M.

how can I send a sketch of the problem for you?
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12/19/22

Doug C.

Here is one way to make a sketch available. Once you have a picture saved to your computer, visit Desmos.com and use the "picture Icon" from the drop-down menu under the upper left hand corner + sign. Here is an example: desmos.com/calculator/ozrjxzaahr You can visit the above link by selecting and right-clicking, choosing "Go to...". Once you have created the graph paste the URL here. You can generate the URL by using the Share Graph Arrow Icon in the upper right hand corner.
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12/19/22

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