Question. 391.
If by = 4x + 5 touches 3y^2 = 8x + 1 at a point P . Find b and coordinates of P . Also find the equation of normal at P .
Answer 391.
Solving by = 4x + 5 with 3y^2 = 8x + 1
=> 3y^2 = 8 * (by - 5)/4 + 1
=> 3y^2 - 2by + 9 = 0
This quadratic shloud have equal roots for the line to be tangent to the parabola
=> discriminant = 0
=> 4b^2 - 108 = 0
=> b = 3√3
y = (1/6) * 2b = √3 and
x = (1/4) (by - 5) = (1/4) * 4 = 1
=> P = (1, √3)
Slope of the line is 4/b = 4/(3√3)
=> slope of the normal is - (3√3)/4
=> equation of the normal through P is
y - √3 = - (3√3)/4 (x - 1)
=> (3√3) x + 4y = 7√3
This answer is verified using the following link.
Wolfram Alpha
Link to YA!
If by = 4x + 5 touches 3y^2 = 8x + 1 at a point P . Find b and coordinates of P . Also find the equation of normal at P .
Answer 391.
Solving by = 4x + 5 with 3y^2 = 8x + 1
=> 3y^2 = 8 * (by - 5)/4 + 1
=> 3y^2 - 2by + 9 = 0
This quadratic shloud have equal roots for the line to be tangent to the parabola
=> discriminant = 0
=> 4b^2 - 108 = 0
=> b = 3√3
y = (1/6) * 2b = √3 and
x = (1/4) (by - 5) = (1/4) * 4 = 1
=> P = (1, √3)
Slope of the line is 4/b = 4/(3√3)
=> slope of the normal is - (3√3)/4
=> equation of the normal through P is
y - √3 = - (3√3)/4 (x - 1)
=> (3√3) x + 4y = 7√3
This answer is verified using the following link.
Wolfram Alpha
Link to YA!
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