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if a line 3x + 4y= -7 1/2 touches the ellipse 3x 2 +4y 2 =1 then the point of contact is, if a line 3x + 4y= -71/2 touches the ellipse 3x2 +4y2 =1 then the point of contact is,
Dear student 3x + 4y= -7^1/2 is a line .......13x2 +4y2 =1 is ellipse .........2Please confirm that it can not be -7^0.5 , It will become imaginary . Please post the image.We will help you Good Luck
Dear studentObtain equation of line in terms of one variable ie x = …Or y = ...Substitute it in second eqnSolve the modified second eqnRoots will be ( -(7)^(-0.5) , -(7)^(-0.5))Which can otherwise be also guessed or verifiedRegards
let line touch ellipse at P(cosp/sqrt3, sinp/2).eqn of ellipse at P would be 3xx1+4yy1= 1 or 3xcosp/sqrt3 + 4ysinp/2= 1or sqrt3xcosp + 2ysinp= 1given line 3x + 4y= -71/2 since both must be identical, we havesqrt3*cosp/3= 2sinp/4= 1/-71/2 or cosp/sqrt3= sinp/2= 1/-71/2 hence P(cosp/sqrt3, sinp/2)= ( – 1/71/2, – 1/71/2).
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