Question

If focal chord of an ellipse y^2=b^2[1-x^2] cuts the ellipse at points whose eccentric angles are 5pi/12 and 23pi/12 then the value of b(b<1) is

Answer 1

Answer:

if focal chord of an ellipse y^ 2=b^2[1-x^2] cuts the ellipse at points whose eccentric angles are 5pi/12

Step-by-step explanation:

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Answer 2

Answer:

The value of $b(b < 1)$ is $-\sqrt{\frac{\cos ^{2} \frac{5 \pi}{12}}{4}+1}$.

Step-by-step explanation:

The focal chord of an ellipse $=y^{2}=b^{2}\left(1-x^{2}\right)$.

$\quad y^{2}=b^{2}\left(1-x^{2}\right)\\&\Rightarrow y^{2}=b^{2}-b^{2} x^{2}\\&\Rightarrow b^{2} x^{2}+y^{2}=b^{2}\\&\Rightarrow \frac{x^{2}}{1}+\frac{y^{2}}{b^{2}}=1\end{aligned}$

We have $a=1$.

The focal chord of the given ellipse cuts the ellipse at points whose eccentric angles are $\frac{5\pi}{12}$ and $\frac{23\pi}{12}$.

Point $p(A \cos \theta, B \sin \theta)$

$A\left[\cos \left(\frac{5 \pi}{12}\right), b \sin \left(\frac{5 \pi}{12}\right)\right]$

And $B\left[\cos \left(\frac{25 \pi}{12}\right), b \sin \left(\frac{23 \pi}{12}\right)\right]$

$B\left[\cos \left(x+\frac{5 \pi}{12}\right), b \sin \left(x+\frac{5 \pi}{12}\right)\right.$

$\Rightarrow B\left[-\cos \frac{5 \pi}{12},-{bsin} \frac{5 \pi}{\sqrt{2}}\right]$

Equation of AB:

$y-\left(-b \sin \frac{5 \pi}{12}\right)=\frac{b \sin \frac{5 \pi}{12}-\left(-b \sin \frac{5 \pi}{12}\right)}{\cos \frac{5 \pi}{12}\left(-\cos \frac{5 \pi}{12}\right)}\left(x-(-b \sin \frac{5 \pi}{12})\right)$ ...... (1)

It passes through $\pi$ at the locus $( ae, 0 )=(e, 0)=\left(\sqrt{1-b^{2}}, 0\right)$.

$\begin{aligned}&b^{2}=a^{2}\left(1-e^{2}\right) \\&b^{2}=1-e^{2} \\&e=\sqrt{1-b^{2}}\end{aligned}$

Putting $e=\sqrt{1-b^{2}}$ in equation 1:

$b \sin \frac{5 \pi}{12}=2 b \tan \frac{5 \pi}{12}\left[\left(\sqrt{1-b^{2}}+\cos \frac{5 \pi}{12}\right]\right.$

$\begin{aligned}& \Rightarrow 1=\frac{2}{\cos \frac{5 \pi}{12}}\left[\sqrt{1-b^{2}}+\cos \frac{5 \pi}{12}\right] \\\Rightarrow & \cos \frac{5 \pi}{12}=2 \sqrt{1-b^{2}}+2 \cos \frac{5 \pi}{12} \\\Rightarrow &-\cos \frac{5 \pi}{12}=2 \sqrt{1-b^{2}}\end{aligned}$

Now let's simplify and find the expression for $b$:

$\frac{-\cos \frac{5 \pi}{12}}{2}= \sqrt{1-b^{2}}\\\frac{-\cos^2 \frac{5 \pi}{12}}{4}-1=-b^2\\b^2=\frac{\cos ^{2} \frac{5 \pi}{12}}{4}+1\\b(b > 1)=\sqrt{\frac{\cos ^{2} \frac{5 \pi}{12}}{4}+1}$, $b(b < 1)=-\sqrt{\frac{\cos ^{2} \frac{5 \pi}{12}}{4}+1}$

Therefore, the value of $b(b < 1)$ is $-\sqrt{\frac{\cos ^{2} \frac{5 \pi}{12}}{4}+1}$.