Question:
<p>Let the ellipse <span class="math-tex">\(3 x^{2}+p y^{2}=4\)</span> pass through the centre <span class="math-tex">\(C\)</span> of the circle <span class="math-tex">\(x^{2}+y^{2}-2 x-4 y-11=0\)</span> of radius <span class="math-tex">\(r\)</span>. Let <span class="math-tex">\(f_{1}, f_{2}\)</span> be the focal distances of the point C on the ellipse. Then <span class="math-tex">\(6 f_{1} {f}_{2}-r\)</span> is equal to</p>
<p style="display:inline">70</p>
<p style="display:inline">68</p>
<p style="display:inline">74</p>
<p style="display:inline">78</p>
Step-by-Step Solution
Key Concept: For a point on an ellipse with semi-major axis a and semi-minor axis b, the sum of focal distances equals 2a (focal radii property). Use the focal distance formula f₁·f₂ = b² + (e²x₀²) where e is eccentricity and (x₀,y₀) is the point on ellipse.
<p>The ellipse equation is given as:<br />
<span class="math-tex">$E: \frac{x^{2}}{4 / 3}+\frac{y^{2}}{4 / P}=1$</span><br />
A circle with center at <span class="math-tex">$(1,2)$</span> has radius:<br />
<span class="math-tex">$r=\sqrt{1+4+11}=4$</span><br />
Since the ellipse passes through the circle's center <span class="math-tex">$(1,2)$</span>,<br />
<span class="math-tex">$\frac{1}{4 / 3}+\frac{4}{4 / P}=1$</span><br />
<span class="math-tex">$P=\frac{1}{4}$</span><br />
The eccentricity <span class="math-tex">$e$</span> is calculated as<br />
<span class="math-tex">$e=\sqrt{1-\frac{4 / 3}{16}}=\sqrt{\frac{11}{12}}$</span><br />
For <span class="math-tex">$a$</span> point <span class="math-tex">$C(h, k)$</span> on the ellipse, the focal distances are:<br />
<span class="math-tex">$F_{1}=4+e \cdot k=4+\sqrt{\frac{11}{12}} \times 2$</span><br />
<span class="math-tex">$F_{2}=4-e \cdot k=4-\sqrt{\frac{11}{12}} \times 2$</span><br />
<span class="math-tex">$\therefore F_{1} F_{2}=16-\frac{11}{3}=\frac{37}{3}$</span><br />
<span class="math-tex">$6 F_{1} F_{2}-r=6 \times \frac{37}{3}-4=70$</span></p>
Correct Answer: A