JEE Advanced · Mathematics · 9. Straight Lines
Paragraph:
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
- A
\(\left(5, \frac{8}{7}\right)\)
- B
\(\left(\frac{7}{5}, \frac{25}{8}\right)\)
- C
\(\left(\frac{11}{5}, \frac{8}{5}\right)\)
- D
\(\left(\frac{8}{25}, \frac{7}{5}\right)\)
Answer & Solution
Correct Answer
(C)
\(\left(\frac{11}{5}, \frac{8}{5}\right)\)
Step-by-step Solution
Detailed explanation
\[
\text { Equation of } A B \text { is }
\]
\[
\begin{aligned}
y-0 & =\frac{\frac{8}{5}}{-\frac{9}{5}-3}(x-3)=\frac{8}{-24}(x-3) \\
\Rightarrow \quad y & =-\frac{1}{3}(x-3) \\
\Rightarrow \quad x & +3 y=3
\end{aligned}
\]
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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