Question

The equation(s) of the standard ellipse passing through the point (4,-1) and touching the line x + 4y - 10 = 0 is/are:

• A. $\frac{x^2}{20}+\frac{y^2}{5}=1$
• B. $\frac{x^2}{5}+\frac{y^2}{20}=1$
• C. $\frac{x^2}{80}+\frac{4y^2}{5}=1$
• D. $\frac{x^2}{5}+\frac{y^2}{80}=1$

Answer 1

Answer: (A), (C)

$\frac{x^2}{80}+\frac{4y^2}{5}=1$

$\mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}a\mathrm{standard}\mathrm{ellipse}\mathrm{be}$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\mathrm{It}\mathrm{passes}\mathrm{through}\left(4,-1\right).\mathrm{So},a^2+16b^2=a^2{b}^2...\left(i\right)\mathrm{Also},x+4y-10=0\mathrm{touches}\mathrm{the}\mathrm{ellipse}.\therefore y=\frac{-x}{4}+\frac{5}{2}\mathrm{comparing}\mathrm{it}\mathrm{with}\mathrm{the}y=\mathrm{mx}\pm \sqrt{a^2{m}^2+b^2}\mathrm{we}\mathrm{have}m=-\frac{1}{4}\mathrm{and}{a}^2{m}^2+b^2=\frac{25}{4}\Rightarrow \frac{25}{4}=\frac{a^2}{16}+b^2\Rightarrow a^2+16b^2=100...\left(\mathrm{ii}\right)\mathrm{From}\left(i\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}{a}^2{b}^2=100\Rightarrow \mathrm{ab}=10$$\mathrm{On}\mathrm{solving}\left(i\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{have}-a=4\sqrt{5},b=\frac{\sqrt{5}}{2}\mathrm{or}a=2\sqrt{5},b=\sqrt{5}\mathrm{Hence}\mathrm{there}\mathrm{are}2\mathrm{ellipses}\mathrm{satisfying}\mathrm{the}\mathrm{given}\mathrm{conditions},i.e.\frac{x^2}{80}+\frac{4y^2}{5}=1\mathrm{and}\frac{x^2}{20}+\frac{y^2}{5}=1$