Question

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. Then length of segment AB is

• A. \(\frac{4\sqrt{2}}{5}\)
• B. \(\frac{16\sqrt{2}}{5}\)
• C. \(\frac{8\sqrt{10}}{5}\)
• D. \(\frac{9\sqrt{10}}{5}\)

Hint:

Write the chord of contact from P, find where it meets the ellipse, then use the distance formula.

Answer 1

The Correct Option is C

Step 1: Chord of contact
The chord of contact is $x+3y=3$.

Step 2: Intersection points
Solving with the ellipse gives the contact points $(3,0)$ and $\left(-\dfrac95,\dfrac85\right)$.

Step 3: Distance
The differences are $\dfrac{24}{5}$ and $\dfrac85$, so $AB=\dfrac85\sqrt{9+1}=\dfrac{8\sqrt{10}}5$, matching the third option.

Final Answer:
The segment AB has length 8 root 10 over 5. \[ \boxed{\text{(C) }\dfrac{8\sqrt{10}}{5}} \]