JEE Advanced · Mathematics · 14. Ellipse
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 equation of the locus of the point whose distance from the point \(P\) and the line \(A B\) are equal, is
- A
\(9 x^2+y^2-6 x y-54 x-62 y\)
\[
+241=0
\]
- B
\(x^2+9 y^2+6 x y-54 x+62 y\)
\[
-241=0
\]
- C
\(9 x^2+9 y^2-6 x y-54 x-62 y\)
\[
-241=0
\]
- D
\(x^2+y^2-2 x y+27 x+31 y\)
\[
-120=0
\]
Answer & Solution
Correct Answer
(A)
\(9 x^2+y^2-6 x y-54 x-62 y\)
\[
+241=0
\]
Step-by-step Solution
Detailed explanation
Equation of \(A B\) is \(y-0=-\frac{1}{3}(x-3)\)
\[
\begin{gathered}
x+3 y-3=0 \\
|x+3 y-3|^2=10\left[(x-3)^2+(y-4)^2\right]
\end{gathered}
\]
(Look at coefficient of \(x^2\) and \(y^2\) in the
answers)
\[
\begin{gathered}
x+3 y-3=0 \\
|x+3 y-3|^2=10\left[(x-3)^2+(y-4)^2\right]
\end{gathered}
\]
(Look at coefficient of \(x^2\) and \(y^2\) in the
answers)
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