mathematics-class-11-unit-12-chapter-22-conic-section-l-9-10-wfota9lhfxo

<h3 id="successexamplesuccess-2"><Success>Example</Success></h3>
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<p>Find the locus of the foot of perpendicular from the foci of an ellipse to the tangents.</p>
<p><strong>Solution:</strong> Consider the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2}$ = 1 , a > b.</p>
<p>Recall that the equation of tangent where slope is m is given by y = mx + c, where $c^2 = a^2 m^2 + b^2$</p>
<p>i.e y = mx $\pm \sqrt {a^2 m^2 + b^2}$</p>
<p>If (h,k) is the foot of perpendicular from foci ($\pm $ ae,0) , then the slope of line joining (h,k) to foci is
k - 0 = $\frac {-1}{m}(h \pm ae)$</p>
<p>$\Rightarrow (mk + h)^2 = a^2 e^2 \longrightarrow $ (1)</p>
<p>Also, (h,k) lies on the tangent line</p>
<p>$\therefore (k-mh)^2 = a^2 m^2 + b^2 \rightarrow $(2)</p>
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<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
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<p>Adding (1)&(2) $\Rightarrow m^2 k^2 + h^2 + 2mhk + k^2 + m^2 h^2 - 2mhk$</p>
<p>=$a^2 e^2 + a^2 m^2 + b^2$</p>
<p>$\Rightarrow (1 + m^2) (h^2 + k^2) = a^2 (1- \frac {b^2}{a^2}) + a^2 m^2 + b^2$</p>
<p>= $a^2 - b^2 + a^2 m^2 + b^2$</p>
<p>= $a^2 (1+m^2)$</p>
<p>$\Rightarrow h^2 + k^2 = a^2$</p>
<p>$\therefore$ The locus is $ x^2 + y^2 = a^2$</p>
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<h3 id="successsolutionsuccess-3"><Success>Solution</Success></h3>
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<p>$\therefore$ Area of $\triangle A B C=\frac{1}{2}\left|\begin{array}{ccc}
a \cos \theta & b \sin \theta & 1 \\a \cos \varphi & b \sin \varphi & 1 \\a \cos \psi & b \sin \psi & 1\end{array}\right|$</p>
<p>=$\frac{1}{2} a b\left|\begin{array}{ccc}\cos \theta & \sin \theta & 1 \\cos \varphi & \sin \varphi & 1 \\cos \psi & \sin \psi & 1
\end{array}\right|$</p>
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<h3 id="successsolutionsuccess-4"><Success>Solution</Success></h3>
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<p>Also, area of $\triangle$ P Q R =$\frac{1}{2}\left|\begin{array}{lll}a \cos \theta & a \sin \theta & 1 \\a \cos \varphi & a \sin \varphi & 1 \\a \cos \psi & a \sin \psi & 1\end{array}\right| $</p>
<p>=$\frac{1}{2} a^2\left|\begin{array}{ccc}\cos \theta & \sin \theta & 1 \\cos \varphi & \sin \varphi & 1 \\cos \psi & \sin \psi & 1
\end{array}\right|$</p>
<p>$\therefore$ $\frac {\text {Area of} \triangle A B C} {\text {Area of} \triangle PQR} =\frac{6}{a}$</p>
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