Question

If the line \(4x-3y+k=0\) touches the ellipse \(5x^{2}+9y^{2}=45\), then the value of \(k\) is

• A. \(21\)
• B. \(\pm 3\sqrt{21}\)
• C. \(\pm 3\)
• D. \(63\)

Hint:

Write the ellipse in standard form and use the tangent condition c squared = a squared m squared + b squared.

Answer 1

The Correct Option is B

Step 1: Substitute the line into the ellipse
From the line, $y=\dfrac{4x+k}{3}$.
Then $5x^{2}+(4x+k)^{2}=45$, i.e. $21x^{2}+8kx+k^{2}-45=0$.

Step 2: Tangent means discriminant zero
$64k^{2}-4\times 21(k^{2}-45)=0$
$-20k^{2}+3780=0$, so $k^{2}=189$.

Step 3: Result
$k=\pm 3\sqrt{21}$.

Final Answer:
A zero discriminant gives k squared = 189, so k = plus or minus 3 root 21. \[ \boxed{\text{(B) }\pm 3\sqrt{21}} \]