MHT CET · Maths · Ellipse
If \(4 x-3 y+k=0\) touches the ellipse \(5 x^{2}+9 y^{2}=45\), then \(k\) is equal to
- A \(\pm 3 \sqrt{21}\)
- B \(3 \sqrt{21}\)
- C \(-3 \sqrt{21}\)
- D \(2 \sqrt{21}\)
Answer & Solution
Correct Answer
(A) \(\pm 3 \sqrt{21}\)
Step-by-step Solution
Detailed explanation
If line \(4 x-3 y+k=0\) touches the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\), then
\(\frac{k}{3}=\sqrt{9 \times\left(\frac{4}{3}\right)^{2}+5}=\pm \sqrt{21}\)
\(\Rightarrow\)
\(k=\pm 3 \sqrt{21}\)
\(\frac{k}{3}=\sqrt{9 \times\left(\frac{4}{3}\right)^{2}+5}=\pm \sqrt{21}\)
\(\Rightarrow\)
\(k=\pm 3 \sqrt{21}\)
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