Vertical tangents on a tilted ellipse

The equation of a closed curve is given by (nx+y)2+4(nx−y)2=20(nx+y)^2+4(nx-y)^2=20, where nn is a constant. Find

(i) the gradient of the curve at the point (x,y)(x,y), marks

(ii) the coordinates of the points on the curve at which the tangent is parallel to the yy-axis, marks

(iii) the equations of the normal at the points where the curve cuts the xx-axis. marks

[2009/IJC/Prelim/I/11]

Total marks: 10

Solution

(i)

Implicit differentiation

The closed curve (nx+y)2+4(nx−y)2=20(nx+y)^2+4(nx-y)^2=20, where nn is a constant. Find the gradient of the curve at the point (x,y)(x,y).

Hint

  1. Differentiate both sides with respect to xx using the chain rule.

  2. Collect dydx\frac{dy}{dx} and solve.

Answer · (i)

dy/dx=n(3y−5nx)/(5y−3nx)dy/dx=n(3y-5nx)/(5y-3nx).

Closed-curve assumption requires n≠0.