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Original question & working

Vertical tangents on a tilted ellipse · source reference

2009/IJC/Prelim/I/11 · Question pp. 1 · Working pp. 2, 3

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The working and teaching notes are included in the interactive lesson.

The answer list prints only one sign in part (iii); both x-axis intersections require a normal. Closedness supplies the unstated n≠0 condition.

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Full original question, working & notes

Printed question

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), [2]

(ii) the coordinates of the points on the curve at which the tangent is parallel to the yy-axis, [4]

(iii) the equations of the normal at the points where the curve cuts the xx-axis. [4]

[2009/IJC/Prelim/I/11]

Printed working

(i) Let (nx+y)2+4(nx−y)2=20(nx+y)^2+4(nx-y)^2=20 be equation (1). Differentiate with respect to xx:

2(nx+y)(n+dydx)+8(nx−y)(n−dydx)=0,2(nx+y)\left(n+\frac{dy}{dx}\right)+8(nx-y)\left(n-\frac{dy}{dx}\right)=0, dydx=n(3y−5nx)5y−3nx.\frac{dy}{dx}=\frac{n(3y-5nx)}{5y-3nx}.

(ii) For tangent parallel to yy-axis gradient is undefined. Thus 5y−3nx=0⇒y=3nx/55y-3nx=0\Rightarrow y=3nx/5. Substitute into (1):

(nx+3nx5)2+4(nx−3nx5)2=20,\left(nx+\frac{3nx}5\right)^2+4\left(nx-\frac{3nx}5\right)^2=20, 64+1625n2x2=20⇒x=±1n(52).\frac{64+16}{25}n^2x^2=20\Rightarrow x=\pm\frac1n\left(\frac52\right).

When x=52nx=\frac5{2n}, y=32y=\frac32; when x=−52nx=-\frac5{2n}, y=−32y=-\frac32. Points are (52n,32)(\frac5{2n},\frac32) and (−52n,−32)(-\frac5{2n},-\frac32).

(iii) At the xx-axis y=0y=0, (nx)2+4(nx)2=20⇒x=±2/n(nx)^2+4(nx)^2=20\Rightarrow x=\pm2/n.

dydx=−5n2(±2/n)−3n(±2/n)=5n3.\frac{dy}{dx}=\frac{-5n^2(\pm2/n)}{-3n(\pm2/n)}=\frac{5n}3.

Normal gradient is −3/(5n)-3/(5n).

At (2/n,0)(2/n,0), y=−35n(x−2/n)=−35nx+65n2y=-\frac3{5n}(x-2/n)=-\frac3{5n}x+\frac6{5n^2}.

At (−2/n,0)(-2/n,0), y=−35n(x+2/n)=−35nx−65n2y=-\frac3{5n}(x+2/n)=-\frac3{5n}x-\frac6{5n^2}.

Qualification

The closed-curve description requires n≠0n\ne0. If n=0n=0, the equation gives the two horizontal lines y=±2y=\pm2, not a closed curve. The answer key prints only the negative-intercept normal; the detailed solution prints both.

Teacher notes

For the vertical tangent the source writes “dy/dxdy/dx undefined ⇒ 1/(dy/dx)=01/(dy/dx)=0”; interpret this as dx/dy=0dx/dy=0, not division by an undefined value.

Gradient of normal =−1/gradient of tangent=−1/(dy/dx)=-1/\text{gradient of tangent}=-1/(dy/dx) when the tangent has finite nonzero gradient.

Answer key

(i) dydx=n(3y−5nx)5y−3nx\frac{dy}{dx}=\frac{n(3y-5nx)}{5y-3nx}. (ii) (52n,32)(\frac5{2n},\frac32) and (−52n,−32)(-\frac5{2n},-\frac32). (iii) Printed short key: y=−35nx−65n2y=-\frac3{5n}x-\frac6{5n^2}; both normals are given above.