if a line 3x + 4y= -7 1/2 touches the ellipse 3x 2 +4y 2 =1 then the point of contact is,

Question

Vikas TU · Answer

Dear student
3x + 4y= -7^1/2 is a line .......1
3x2 +4y2 =1 is ellipse .........2
Please confirm that it can not be -7^0.5 , It will become imaginary .

Please post the image.

We will help you

Good Luck

Saurabh Koranglekar · Answer

Dear student

Obtain equation of line in terms of one variable ie x = …

Or y = ...

Substitute it in second eqn

Solve the modified second eqn

Roots will be ( -(7)^(-0.5) , -(7)^(-0.5))

Which can otherwise be also guessed or verified

Regards

Aditya Gupta · Answer

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= 1 or sqrt3xcosp + 2ysinp= 1 given line 3x + 4y= -7 1/2 since both must be identical, we have sqrt3*cosp/3= 2sinp/4= 1/ -7 1/2 or c osp/sqrt3= sinp/2= 1/ -7 1/2 hence P(cosp/sqrt3, sinp/2)= ( – 1/ 7 1/2 , – 1/ 7 1/2 ).

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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= -7^1/2 touches the ellipse 3x² +4y² =1 then the point of contact is,

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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,

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if a line 3x + 4y= -7^1/2 touches the ellipse 3x² +4y² =1 then the point of contact is,