An ellipse passes through the point `(4,-1)` and touches the line `x+4y-10=0` . Find its equation if its axes coincide with the coordinate axes.

Explanation

Let the equation of the ellipse be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1" "(1)`
Since it passes through the point (4,-1), we have `a^(2)+16b^(2)=a^(2)b^(2) " "(2)`
Equation of given tangent is
`x+4y-10=0" "(2)`
`y=-(1)/(4)x+(5)/(2)`
Comparing with `y=mx+c`, we have `m=-(1)/(4),c=(5)/(2)` Now, `c^(2)=a^(2)m^(2)+b^(2)`
`implies(25)/(4)=a^(2)/(16)+b^(2)`
`impliesa^(2)+16b^(2)=100" "(3)`
From (2) and (3), we get
`a^(2)b^(2)=100orab=10" "(4)`
Solving (3) and (4), we get
`4sqrt5andb=(sqrt5)/(2)ora=2sqrt5,b=sqrt5`
Hence, the equations of ellipses are
`(x^(2))/(80)+(4y)/(5)=1and(x^(2))/(20)+(y^(2))/(5)=1`