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Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The orthocentre of the triangle $P A B$ is
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Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The orthocentre of the triangle $P A B$ is
Solution:
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Verified Answer
The correct answer is:
$\left(\frac{11}{5}, \frac{8}{5}\right)$
$\left(\frac{11}{5}, \frac{8}{5}\right)$
$$
\text { Equation of } A B \text { is }
$$
$$
\begin{aligned}
y-0 & =\frac{\frac{8}{5}}{-\frac{9}{5}-3}(x-3)=\frac{8}{-24}(x-3) \\
\Rightarrow \quad y & =-\frac{1}{3}(x-3) \\
\Rightarrow \quad x & +3 y=3
\end{aligned}
$$
Equation of the straight line perpendicular to $A B$ through $P$ is $3 x-y=5$.
Equation of $P A$ is $x-3=0$.
The equation of straight line perpendicular to $P A$ through $B\left(\frac{-9}{5}, \frac{8}{5}\right)$. is $y=\frac{8}{5}$.
Hence, the orthocentre is $\left(\frac{11}{5}, \frac{8}{5}\right)$.
\text { Equation of } A B \text { is }
$$
$$
\begin{aligned}
y-0 & =\frac{\frac{8}{5}}{-\frac{9}{5}-3}(x-3)=\frac{8}{-24}(x-3) \\
\Rightarrow \quad y & =-\frac{1}{3}(x-3) \\
\Rightarrow \quad x & +3 y=3
\end{aligned}
$$
Equation of the straight line perpendicular to $A B$ through $P$ is $3 x-y=5$.
Equation of $P A$ is $x-3=0$.
The equation of straight line perpendicular to $P A$ through $B\left(\frac{-9}{5}, \frac{8}{5}\right)$. is $y=\frac{8}{5}$.
Hence, the orthocentre is $\left(\frac{11}{5}, \frac{8}{5}\right)$.
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