Question

The equation of the tangent to the ellipse x^2+16y^2=16 making an angle of 60 wit x axis is

Answer 1

Answer:

7√3 x - 7 y + 49 = 0

and parallel to this, on the other side of the ellipse,

7√3 x - 7 y - 49 = 0.

Step-by-step explanation:

Differentiating the equation of the ellipse, we have

2x dx + 32 y dy = 0

=> dy / dx = - x / (16y).

For the tangent to make an angle of 60 degrees with the x-axis, we need the gradient to be tan(60 deg) = √3 (I'm assuming we mean the angle made on the right hand side of the intersection; to include the case where the angle on the left is 60 degrees, we need to change this to -√3).

So a point on the ellipse has a tangent of this gradient if

dy / dx = √3

or in other words, if

x = -16 √3 y.

Since this is a point of the ellipse, it satisfies the equation of the ellipse, so we put this into there to find

x² + 16 y² = 16

=> (16²) (3) y² + 16 y² = 16

=> 48 y² + y² = 1

=> y = ±1 / 7.

From above, the corresponding values of x are ±( -16 √3 / 7 ).

Now for one tangent, we have the point ( 1/7, -16√3/7) and the gradient √3, so

( y - 1/7 ) / ( x + 16√3 / 7 ) = √3

which rearranged is

7√3 x - 7 y + 49 = 0.

For the other tangent, we have the point ( -1/7, 16√3/7) and the gradient √3, so

( y + 1/7 ) / ( x - 16√3 / 7 ) = √3

which rearranged is

7√3 x - 7 y - 49 = 0.

P.S. If you also need the ones that make an angle of 60 degrees on the left, just use the symmetry of ellipse to notice that intercepts are the same while the gradient changes sign, so the equations are obtained just by changing the sign of the y terms:

7√3 x + 7 y + 49 = 0

and

7√3 x + 7 y - 49 = 0.