Question

Show that the line y=mx+c touches the elipse x^2/a^2+y^2/b^2=1 if c^2=a^2m^2+b^2

Answer 1

Given----> Line y = mx + c touches the ellipse

x² / a² + y² / b² = 1 .

To prove ----> If line touches the ellipse then

c² = a² m² + b²

Concept used ---->

1) Any line touches any curve if it intersect line at one point only .

2) If roots of quadratic equation ax² + bx + c = 0 is equal than

b² - 4ac = 0

Solution----> ATQ,

Equation of ellipse is

x² / a² + y² / b² = 1

Multiplying whole equation by a²b² , we get,

a²b² ( x²/a² ) + a²b² ( y²/b² ) = a²b²

b² x² + a² y² = a²b² ...................( 1 )

Now equation of line ,

y = mx + c .......................( 2 )

Putting y = mx + c in equation ( 1 )

=> b²x² + a² ( mx + c )² = a²b²

=> b²x² + a² ( m²x² + c² + 2cmx ) = a²b²

=> b²x² + a²m²x² + a²c² + 2cma²x - a²b² = 0

=> ( b² + a²m² ) x² + 2cma² x + ( a²c² - a²b² ) = 0

Now, comparing this equation with,

Ax² + Bx + C = 0

A= ( b² + a²m² ) , B = 2a²cm , C = ( a²c² - a²b² )

Now , line ( 2 ) touches the ellipse ( 1 ) , so above equation has same root , so ,

B² - 4 AC = 0

=> ( 2a²cm )² - 4 ( b² + a²m² ) ( a²c² - a²b² ) = 0

=> 4a⁴c²m² = 4 ( b² + a²m² ) ( a²c² - a²b² )

=> 4a⁴c²m² = 4a² ( b² + a²m² ) ( c² - b² )

4a² is cancel out from each side

=> a²m²c² = ( b² + a²m² ) ( c² - b² )

=> a²m²c² = b²c² - b⁴ + a²m²c² - a²m²b²

a²m²c² is cancel out from each side and we get,

=> 0 = b²c² - b⁴ - a²m²b²

=> b⁴ + a²m²b² = b²c²

=> b² ( b² + a²m² ) = b²c²

b² is cancel out from each side and we get,

=> b² + a²m² = c²

=> c² = a² m² + b²

Which is required condition .

For second method plzz see the attachement