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Find the locus of point of intersection of lines $ y + mx = \sqrt{a^2m^2 + b^2}$ and $ my - x = \sqrt{a^2 + m^2b^2}$

The point of intersection say, $(h,k)$ must satisfy both the equation. When I tried solving both the equation it didn't help me. What am I doing wrong?

Kaushik
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2 Answers2

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Here is what you can do:

Square both equations

$$ y^2 + 2mxy +m^2 x^2= a^2m^2 + b^2$$ $$ m^2y^2 - 2mxy +x^2= a^2 + m^2b^2$$

Add them up to get a circle,

$$y^2+x^2=a^2+b^2$$

Quanto
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There is a perfect answer up here by Quanto so just a hint: The equations bear similarity to the standard (slope-form) equation of a tangent to an ellipse.
You can notice by adjusting the equations that they are perpendicular to each other. Which means, that from a point outside the ellipse you are drawing 2 perpendicular tangents to it. By definition, the locus of such an intersecting point is called a Director Circle.

The locus would hence have to be in the form of an equation of a circle.

ali
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