Question

Find equations for two lines through the origin that are tangent to the ellipse
7x2−168x+y2+62=0
please show me the steps :)

Answer 1

Rewrite the ellipse as:
7(x-12)²+y²-946=0
So it is an ellipse on the x-axis.

Check that the origin is exterior to the ellipse: set y=0,
x-12=±sqrt(946/7)=±11.625
x=0.375 or 23.625
=> the origin is exterior to the ellipse.

Slope of a point on the ellipse can be found by implicit differentiation of the ellipse equation:
14x-168+2y(dy/dx)=0 from which
dy/dx = (168-14x)/2y
Since the tangent line passes through the orgin, at the point of tangency, dy/dx=y/x, or
(168-14x)/2y = y/x
Transpose and solve for y:
y^2=84x-7x^2
Substitute into the ellipse equation to get y:
7x^2-168x+62+84x-7x^2=0
Solve for x:
84x=62 => x=31/42
Calculate y:
y=sqrt(84x-7x^2)
=sqrt(84*31/42-7(31/42)^2)
=sqrt(14663)/(6*sqrt(7))
=7.628 approx.

Therefore the tangent line is given by:
y=[sqrt(14663)/(6*sqrt(7))]/(31/42)x
=[sqrt(7)*sqrt(14663)/31]x

Check by graphing and check that the point is on the line.

Answer 2

thanks :)