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Question: Answered & Verified by Expert
The locus of the point of intersection of perpendicular tangents to the ellipse is called
MathematicsEllipseKCETKCET 2012
Options:
  • A hyperbola
  • B ellipse
  • C auxiliary circle
  • D director circle
Solution:
1066 Upvotes Verified Answer
The correct answer is: director circle
The equation of a pair of tangents from $(\alpha, \beta)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is
$$
\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right)=\left(\frac{x \alpha}{a^{2}}+\frac{y \beta}{b^{2}}-1\right)^{2}
$$
$\left(\mathrm{SS}_{1}=\mathrm{T}^{2}\right)$
The tangents are perpendicular, if the coefficient of $x^{2}+$ the coefficient of $y^{2}=0$.
$$
\begin{array}{r}
\Rightarrow \frac{1}{a^{2}}\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right)-\frac{\alpha^{2}}{a^{4}}+\frac{1}{b^{2}}\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) \\
-\frac{\beta^{2}}{b^{4}}=0
\end{array}
$$
$$
\begin{gathered}
\Rightarrow \quad \frac{\beta^{2}}{a^{2} b^{2}}-\frac{1}{a^{2}}+\frac{\alpha^{2}}{a^{2} b^{2}}-\frac{1}{b^{2}}=0 \\
\Rightarrow \quad \alpha^{2}+\beta^{2}=a^{2} b^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \\
=a^{2} b^{2}\left(\frac{a^{2}+b^{2}}{a^{2} b^{2}}\right) \\
=a^{2}+b^{2}
\end{gathered}
$$
Hence, the locus of $(\alpha, \beta)$ is the circle
$$
x^{2}+y^{2}=a^{2}+b^{2}
$$
This circle is called the director circle.

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