Find the equation of the parabola whose latus rectum is 4 units,axis is the line 3x+4y-4=0 and the tangent at the vertex is the line 4x-3y+7=0.

Dear student
Kindly follow the another and easy approach of doing the question.

Let Px,y be any point on the parabola.Draw PN  to its axis and PM  to the tangent at the vertex.Theny2=4axPM2=Latus ractum×PNPM2=Latus rectum×PN  ...1      Because latus rectum of y2=4ax is 4aNow,PM=3x+4y-49+16    as distance of a point from directrix is : 3x+4y-432+42PM=3x+4y-45PN=lenght of  from Px,y on the tangent at the vertex APN=4x-3y+716+9PN=4x-3y+75Latus rectum=4 givenFrom 13x+4y-452=44x-3y+75or 3x+4y-42=204x-3y+7is the required eq of parabola.
Regards

  • 3
What are you looking for?