mathematics-class-11-unit-12-chapter-17-conic-section-l-4-10-cchgpvujj9u

### <Success>Problem</Success>
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<p>Find the equation of the hyperbola whose foci are at $( \pm 4,0)$ and the length of the latex rectum is 12 .</p>
<p><strong>Solution :</strong></p>
<p>Since foci are on the $x$-axis, the Equation is</p>
<p>$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$</p>
<p>Foci $=( \pm 4,0) \Rightarrow c=4 \Rightarrow a^2+b^2=4^2$ (i)</p>
<p>Length of latex rectum, $l=\frac{2 b^2}{a}=12$</p>
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<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>$
\Rightarrow b^2=6 a \text {-(ii) }
$</p>
<p>(i)
$ \Rightarrow b^2=6 a \quad \text { (i) } and (ii)  \Rightarrow a^2+6 a=16\Rightarrow a^2+6 a-16=0 $</p>
<p>$\Rightarrow(a-2)(a+8)=0$</p>
<p>$\Rightarrow \quad a=2 \quad[\because a>0]$</p>
<p>$\therefore \quad b^2=6 a=12$</p>
<p>$\therefore$ The equation is $\frac{x^2}{4}-\frac{y^2}{12}=1$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/495f46d6-94da-4c98-4ce8-63273d7a8b00/public"></p>
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### <Success>Solution</Success>
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<p>putting this in (i), gives</p>
<p>$\frac{9}{a^2}-\frac{4}{10-a^2}=1 $</p>
<p>$\Rightarrow 9\left(10-a^2\right)-4 a^2=a^2\left(10-a^2\right) $</p>
<p>$\Rightarrow 90-9 a^2-4 a^2=10 a^2-a^4 $</p>
<p>$\Rightarrow a^4-23 a^2+90=0 $</p>
<p>$\Rightarrow \left(a^2-5\right)\left(a^2-18\right)=0 $</p>
<p>$\Rightarrow a^2=5 \quad \frac{y^2}{5}-\frac{x^2}{5}=1 $</p>
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<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
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<p>$\text { But } a^2+b^2=10 \Rightarrow a^2=18$</p>
<p>But, $a^2+b^2=10 \Rightarrow a^2=18$ $\frac{y^2}{5}-\frac{x^2}{5}=1$</p>
<p>Hence $a^2+b^2=10 \Rightarrow a^2 \leq 10$</p>
<p>i.e. $y^2-x^2=5$</p>
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### <Success>Asymptotes of a hyperbola</Success>
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<p>The straight lines $y= \pm \frac{b}{a} x$ do not intersect the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ because</p>
<p>$
y= \pm \frac{b}{a} x \Rightarrow \frac{y^2}{b^2}=\frac{x^2}{a^2} \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=0
$</p>
<p>$\Rightarrow(x, y)$ does not lie on the hyperbola.</p>
<p>However, if $(x, y)$ on the hyperbola and $(x, y$) lies on the line $y=\frac{b}{a} x$, then</p>
<p>$
\begin{aligned}
\left|y-y_1\right| & =\left|b \sqrt{\frac{x^2}{a^2}-1}-\frac{b}{a} x\right| \
& =\frac{b}{a}\left(x-\sqrt{x^2-a^2}\right)
\end{aligned}
$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0332b2aa-719d-45f9-7051-620513f78500/public"></p>
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### <Success>Asymptotes of a hyperbola</Success>
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<p>As $x \rightarrow+\infty$,</p>
<p>$
\begin{aligned}
& 1 \lim _{x \rightarrow+\infty}\left[x-\sqrt{x^2-a^2}\right] \\
& =\lim _{x \rightarrow+\infty} \frac{\left(x+\sqrt{x^2-a^2}\right)\left(x-\sqrt{x^2-a-1}\right)}{\left(x+\sqrt{x^2-a}\right)} \\
& =\lim _{x \rightarrow \infty} \frac{x^2-\left(x^2-a^2\right)}{x+\sqrt{x^2-a^2}} \\
& =\lim _{x \rightarrow+\infty} \frac{a^2}{x+\sqrt{x^2-a^2}}=0
\end{aligned}
$</p>
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<h3 id="successasymptotes-of-a-hyperbolasuccess"><Success>Asymptotes of a hyperbola</Success></h3>
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<p>$\therefore$ The line $y=\frac{b}{a} x$ and hyperbola</p>
<p>$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ approaches each other as $x \rightarrow+\infty$.</p>
<p>These lines are therefore the asymptotes of the hyperbola.</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/64702836-49ed-407d-21c9-3977ba43ff00/public"></p>
<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/ec8ae602-b659-4bff-40af-411205415a00/public"></p>
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