Question

An ellipse passes through the point $(4,-1)$ and its axes are along the axes of coordinate. If the line $x+4y-10=0$ is a tangent to it, then its equation is

• A. $\frac{x^2}{100}+\frac{y^2}{5}=1$
• B. $\frac{x^2}{80}+\frac{y^2}{5/4}=1$
• C. $\frac{x^2}{20}+\frac{y^2}{5}=1$
• D. None of these

Answer 1

Answer: (B), (C)

$\frac{x^2}{80}+\frac{y^2}{5/4}=1$
$\frac{x^2}{20}+\frac{y^2}{5}=1$

$Lettheequationofellipsebe:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1Or,{b^{2}x}^2+y^2{a^2=a}^2{b}^2----------\left(1\right)Line:x+4y-10=0Or,x=10-4y-------\left(2\right)Solvingthelineandellipseweget:{b^2(10-4y)}^2+y^2{a^2=a}^2{b}^{2}Or,b^2(100+16y^2-80y)+y^2{a^2=a}^2{b}^{2}Or,y^2(16b^2+a^2)+y(-80b^2)+100b^2-a^2{b}^2=0---------\left(3\right)Sincethelineistangenttotheellipse,SoD=0forthisquadraticequation3\therefore 6400b^4-4(16b^2+a^2)(100b^2-a^2{b}^2)=0Or,1600b^4-(1600b^4-16b^4{a}^2+100a^2{b}^2-{a^{4}b}^2)=0Or,16b^4{a}^2-100a^2{b}^2+{a^{4}b}^2=0Or,16b^2-100+a^2=0----------\left(4\right)And,(4,-1)liesontheellipsethen,{16b}^2+{a^2=a}^2{b}^2--------\left(5\right)Byequation4and5wegetab=\pm 10Or,a=\frac{\pm 10}{b}\therefore {16b}^2+\frac{100}{b^2}=100Onsolvingweget,b^2=5and{b}^2=\frac{5}{4}And,a^2=80a^2=20Equationofellipseare:\frac{x^2}{80}+\frac{y^2}{\frac{5}{4}}=1And,\frac{x^2}{20}+\frac{y^2}{5}=1$

Option [B][C]