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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Ellipse

Let the ellipse 3 x^2+py^2=4 pass through the centre C of the...

Let the ellipse 3x2+py2=43 x^2+\mathrm{py}^2=4 pass through the centre CC of the circle x2+y2−2x−4y−11=0x^2+y^2-2 x-4 y-11=0 of radius rr. Let f1,f2f_1, f_2 be the focal distances of the point C on the ellipse. Then 6f1f2−r6 f_1 f_2-r is equal to

Options

  1. A74
  2. B68
  3. C70
  4. D78

Correct answer

C. 70

Step-by-step solution

E:x24/3+y24/P=1\mathrm{E}: \frac{\mathrm{x}^2}{4 / 3}+\frac{\mathrm{y}^2}{4 / \mathrm{P}}=1Centre of circle ( 1,2 ), radiusr=1+4+11r=4\begin{aligned}& \mathrm{r}=\sqrt{1+4+11} \\ & \mathrm{r}=4\end{aligned}\becauseEpassfromcentreE pass from centre(1,2) \therefore \frac{3}{4}+\mathrm{P}=1 \mathrm{P}=\frac{1}{4} \quad \thereforeverticalellipsevertical ellipsee=1−4/316=1−112=1112∴ Focal distance of C(h,k)=b±ekF1=4+1112×2F2=4−1112×2∴F1F2=16−113=373∴6F1F2−r=74−4=70\begin{aligned} & \mathrm{e}=\sqrt{1-\frac{4 / 3}{16}}=\sqrt{1-\frac{1}{12}}=\sqrt{\frac{11}{12}} \\ & \therefore \text{ Focal distance of } \mathrm{C}(\mathrm{h}, \mathrm{k}) \\ & =\mathrm{b} \pm \mathrm{ek} \\ & \mathrm{F}_1=4+\sqrt{\frac{11}{12}} \times 2 \\ & \mathrm{ F}_2=4-\sqrt{\frac{11}{12}} \times 2 \\ & \therefore \mathrm{ F}_1 \mathrm{ F}_2=16-\frac{11}{3}=\frac{37}{3} \\ & \therefore 6 \mathrm{ F}_1 \mathrm{ F}_2-\mathrm{r}=74-4=70\end{aligned}$
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