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If a parabola passess through the points $(-2,1)$, $(1,2)$ and $(-1,3)$ having horizontal axis, then the length of the latus rectum of that parabola is
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Verified Answer
The correct answer is:
$\frac{2}{5}$
Let the equation of parabola having horizontal axis, vertex at $(h, k)$ is
$$
(y-k)^2=4 a(x-h)
$$
$\because$ Parabola (i) passes through the points $(-2,1)$, $(1,2)$ and $(-1,3)$, so
$$
\begin{aligned}
& (k-1)^2=4 a(-2-h) \\
& \Rightarrow \quad k^2-2 k+1=-8 a-4 a h \\
& \Rightarrow \quad\left(k-2^2=4 a(1-h)\right. \\
& \Rightarrow \quad k^2-4 k+4=4 a-4 a h \\
& \text { and } \quad(3-k)^2=4 a(-1-h) \\
& \Rightarrow \quad k^2-6 k+9=-4 a-4 a h \\
&
\end{aligned}
$$
From Eqs. (ii), (iii) and (iv), we get
$$
4 a=\frac{2}{5}
$$
$$
(y-k)^2=4 a(x-h)
$$
$\because$ Parabola (i) passes through the points $(-2,1)$, $(1,2)$ and $(-1,3)$, so
$$
\begin{aligned}
& (k-1)^2=4 a(-2-h) \\
& \Rightarrow \quad k^2-2 k+1=-8 a-4 a h \\
& \Rightarrow \quad\left(k-2^2=4 a(1-h)\right. \\
& \Rightarrow \quad k^2-4 k+4=4 a-4 a h \\
& \text { and } \quad(3-k)^2=4 a(-1-h) \\
& \Rightarrow \quad k^2-6 k+9=-4 a-4 a h \\
&
\end{aligned}
$$
From Eqs. (ii), (iii) and (iv), we get
$$
4 a=\frac{2}{5}
$$
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