KCET · Maths · Area Under Curves
If the area of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1\) is \(20 \pi\) sq units, then \(\lambda\) is
- A \(\pm 4\)
- B \(\pm 3\)
- C \(\pm 2\)
- D \(\pm 1\)
Answer & Solution
Correct Answer
(A) \(\pm 4\)
Step-by-step Solution
Detailed explanation
Area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(\pi|a b|\).
\(a^{2}=25, b^{2}=\lambda^{2}\)
Area of the ellipse \(\frac{x^{2}}{25}+\frac{b^{2}}{\lambda^{2}}=1\) is \(\pi \times 5 \times|\lambda|\).
\(\begin{array}{llrl}
\Rightarrow & & 20 \pi & =5 \pi|\lambda| \\
\Rightarrow & & |\lambda| & =4 \\
\Rightarrow & & \lambda & =\pm 4
\end{array}\)
\(a^{2}=25, b^{2}=\lambda^{2}\)
Area of the ellipse \(\frac{x^{2}}{25}+\frac{b^{2}}{\lambda^{2}}=1\) is \(\pi \times 5 \times|\lambda|\).
\(\begin{array}{llrl}
\Rightarrow & & 20 \pi & =5 \pi|\lambda| \\
\Rightarrow & & |\lambda| & =4 \\
\Rightarrow & & \lambda & =\pm 4
\end{array}\)
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