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Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular.
Let the equation to the tangent be $$y=mx+\sqrt{a^2m^2+b^2} $$ This has to roots for $m$ that i.e $m_1$ and $m_2$.

perpendicular to this line passing through $(0,0)$ is $$my+x=0 $$ slope is $\frac{-1}{m}$ so for perpendiculars $$\frac{(-1)(-1)}{m_1m_2} =-1 $$ so $m_1m_2=-1 $ from equation of tangent $y=mx+\sqrt{a^2m^2+b^2} $ $$m^2(x^2-a^2)-2mxy+y^2-b^2=0 $$and hence locus is $$\frac{y^2-b^2}{x^2-a^2}=-1 $$ But this is not the correct answer. The answer is $b^4x^2+a^4y^2=a^2b^2(a^2+b^2) $. What's the error ?

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  • $\begingroup$ why you are taking line perpendicular to tangent, its given that lines joining points of contact to origin are perpendicular. $\endgroup$ Oct 16, 2015 at 6:01

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Let $(x_1,y_1)$ be the generic point of the locus.

It is well known that $$\frac {xx_1}{a^2}+\frac {yy_1}{b^2}=1$$ represents the line passing through the points of contact of the tangents from $(x_1,y_1)$.

Now consider the equation $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\left(\frac {xx_1}{a^2}+\frac {yy_1}{b^2}\right)^2=0$$ It is satisfied by the coordinates of the center and the points of contact.

Since it can be written $$x^2 \left(\frac {x_1^2}{a^4}-\frac 1{a^2}\right) + y^2 \left(\frac {y_1^2}{b^4}-\frac 1{b^2}\right) + \frac {2\,x\,y\,x_1y_1}{a^2b^2}=0$$

it is quadratic homogeneous so represents a pair of lines (degenerate conic) , clearly the lines joining the points of contact to the centre.

It is not difficult to prove that the lines are mutually perpendicular iff the sum of the coefficients of $x^2$ and $y^2$ is zero, that is $$\frac {x_1^2}{a^4}+\frac {y_1^2}{b^4}=\frac 1{a^2}+\frac 1{b^2}$$

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  • $\begingroup$ well where have I mistaken ? $\endgroup$ Oct 17, 2015 at 10:00
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    $\begingroup$ The lines joining the points of contact to the centre must be mutually perpendicular not everyone perpendicular to the respective tangent. Essentially you cannot avoid considering the coordinates of the points of contact, but in this way the computation is very heavy. $\endgroup$ Oct 17, 2015 at 13:37

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