Question:

The ellipse x2+4y2=4x^2 + 4y^2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4,0)(4, 0). Then the equation of the ellipse is

Updated On: Aug 15, 2022
  • x2+16y2=16x^2 +16y^2 = 16
  • x2+12y2=16x^2 +12y^2 = 16
  • 4x2+48y2=484x^2 + 48y^2 = 48
  • 4x2+64y2=484x^2 + 64y^2 = 48
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

x2+4y2=4x24+y21=1a=2,b=1P=(2,1)x^{2}+4y^{2}=4 \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{1}=1 \Rightarrow a=2, b=1 \Rightarrow P=\left(2, 1\right) Required Ellipse is x2a2+y2b2=1x242+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{4^{2}}+\frac{y^{2}}{b^{2}}=1 (2,1)\left(2, 1\right) lies on it 416+1b2=11b2=114=34b2=43\Rightarrow \frac{4}{16}+\frac{1}{b^{2}}=1 \Rightarrow \frac{1}{b^{2}}=1-\frac{1}{4}=\frac{3}{4} \Rightarrow b^{2}=\frac{4}{3} x216+y2(43)=1x216+3y24=1x2+12y2=16\therefore \frac{x^{2}}{16}+\frac{y^{2}}{\left(\frac{4}{3}\right)}=1 \Rightarrow \frac{x^{2}}{16}+\frac{3y^{2}}{4}=1 \Rightarrow x^{2}+12y^{2}=16
Was this answer helpful?
0
0

Top Questions on Conic sections

View More Questions