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Find the value of k and hence the point of contact of the tangent line 4x+y+k=0\, with the ellipse x^{2}+3y^{2}=3\,

From the given equations l=4,m=1,n=k,a^{2}=3,b^{2}=1\,

Substituting the values in the condition of tangency,we get


k^{2}=49,k=\pm 7\,

Therefore tangents are 4x+y+7=0,4x+y-7=0\,

The points of contact are

\left({\frac  {(-3)(4)}{7}},{\frac  {(-1)(1)}{7}}\right)\,

\left({\frac  {-12}{7}},{\frac  {-1}{7}}\right)\,

Other point is \left({\frac  {12}{7}},{\frac  {1}{7}}\right)\,

Main Page:Geometry:The Ellipse