mathematics-class-11-unit-12-chapter-15-conic-section-l-2-10-rrn-kxqigqc

### <Success>Formula for standard ellipse</Success>
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<p>$ \Leftrightarrow \quad(x+c)^2+y^2$
$=4 a^2+(x-c)^2+y^2$
$-4 a \sqrt{(x-c)^2+y^2}$</p>
<p>$ \Leftrightarrow 4 a \sqrt{(x-c)^2+y^2}$
$=4 a^2+(x-c)^2-(x+c)^2={-4 x c}$</p>
<p>$ \Leftrightarrow 4 a \sqrt{(x-c)^2+y^2}=4(a^2-c x)$</p>
<p>$ \Leftrightarrow a^2[(x-c)^2+y^2]=(a^2-c x)^2$
$=a^4-2 c a^2 x +c^2x^2$</p>
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<h3 id="successformula-for-standard-ellipsesuccess"><Success>Formula for standard ellipse</Success></h3>
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<p>$\Leftrightarrow a^2 x^2+a^2 c^2-2 a^2 c x+a^2 y^2$
$=a^4-2 a^2 cx+c^2 x^2 $</p>
<p>$\Leftrightarrow (a^2-c^2)= x^2+a^2 y^2$
$=a^4-a^2 c^2=a^2 (a^2-c^2)$</p>
<p>$\Leftrightarrow b^2 x^2+a^2 y^2=a^2 b^2$</p>
<p>$\Leftrightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $</p>
<p>[Symmetric about both x and y axis]</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/98f6e62c-5142-48d9-2a23-39eb59f3f200/public"></p>
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### <Success>Latus rectum</Success>
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<p>Line segment with endpoints on the ellipse passing through a focus and perpendicular to the major axis.</p>
<p>Let $P F=Q F=l$.</p>
<p>He need to find $2 l$.</p>
<p>$
P \equiv(c, l)=(a e, l)
$</p>
<p>Since $P(a e, l)$ lies on the ellipse</p>
<p>$\frac{x^2}{a^2}+\frac{l^2}{b^2}=1$,</p>
<p>$\frac{a^2 e^2}{a^2}+\frac{l^2}{b^2}=1 $</p>
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<h3 id="successlength-of-latus-rectumsuccess"><Success>Length of latus rectum</Success></h3>
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<p>$\Rightarrow \frac{l^2}{b^2}=1-e^2=1-\frac{c^2}{a^2}$</p>
<p>$ \Rightarrow \frac{l^2}{b^2}=\frac{a^2-c^2}{a^2}=\frac{b^2}{a^2}$</p>
<p>$\Rightarrow l^2=\frac{b^4}{a^2} $</p>
<p>$ \Rightarrow l=\frac{b^2}{a}$</p>
<p>$$\Rightarrow 2 l=\frac{2 b^2}{a}=\text { length of latus rectum }$$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b7215470-3396-4ed3-3fcb-2d1cabef5a00/public"></p>
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