The locus of the point of intersection of per | vedclass

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Question

(c) Let point be $(h,k)$ their pair of tangent will be

$\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 1} \right)\,\left( {\frac{{{h^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Let point be $(h,k)$ their pair of tangent will be

$\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 1} \right)\,\left( {\frac{{{h^2}}}{{{a^2}}} + \frac{{{k^2}}}{{{b^2}}} - 1} \right) = {\left( {\frac{{hx}}{{{a^2}}} + \frac{{yk}}{{{b^2}}} - 1} \right)^2}$

Pair of tangents will be perpendicular, if

coefficient of ${x^2}$ + coefficient of ${y^2} = 0$

==>$\frac{{{k^2}}}{{{a^2}{b^2}}} + \frac{{{h^2}}}{{{a^2}{b^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$

==> ${h^2} + {k^2} = {a^2} + {b^2}$

Replace $(h,k)$ by $(x,y)$

==> ${x^2} + {y^2} = {a^2} + {b^2}$.

The locus of the point of intersection of perpendicular tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is

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