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A particle is travelling in clockwise direction on the ellipse $\frac{x^2}{100}+\frac{y^2}{25}=1$. If the particle leaves the ellipse at the point $(-8,3)$ on it and travels along the tangents to the ellipse at that point then the point where the particle crosses the $\mathrm{Y}$-axis is
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Verified Answer
The correct answer is:
$\left(0, \frac{25}{3}\right)$
$\frac{x^2}{100}+\frac{y^2}{25}=1$
Equation of tangent at $(-8,3)$
$\begin{aligned}
& \frac{-8 x}{100}+\frac{3 y}{25}=1 \Rightarrow-8 x+12 y=100 \\
\Rightarrow \quad & 2 x-3 y+25=0
\end{aligned}$
For point at $y$ axis, $x=0$
$\therefore \quad-3 y+25=0 \Rightarrow y=\frac{25}{3}$
$\therefore \quad$ Required point is $\left(0, \frac{25}{3}\right)$.
Equation of tangent at $(-8,3)$
$\begin{aligned}
& \frac{-8 x}{100}+\frac{3 y}{25}=1 \Rightarrow-8 x+12 y=100 \\
\Rightarrow \quad & 2 x-3 y+25=0
\end{aligned}$
For point at $y$ axis, $x=0$
$\therefore \quad-3 y+25=0 \Rightarrow y=\frac{25}{3}$
$\therefore \quad$ Required point is $\left(0, \frac{25}{3}\right)$.
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