Ellipse tangent intercepts and their midpoint

C

Total marks: 10

Solution

(a)

Connected rates of change

$Q$ moves on $C$ so that its $x$ -coordinate increases at 15 units per second. Find the exact rate of change of $\theta$ when $Q=\left(2,\frac{3\sqrt3}2\right)$ .

Explore this idea

Hint

Find $\theta$ at $Q$ from $x=4\cos\theta=2$ .
$\frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot\frac{dx}{dt}$ .

Answer · (a)

d\theta/dt=-5\sqrt3/2

Explore the idea

Q moves along the ellipse

As Q moves right (x increasing), θ decreases from π towards 0.

Position along the curve(2, 2.59808)θ = 1.0472

Try an example

Tangent gradient

-0.43301

Normal gradient

2.3094

At $Q$ : $\theta=\frac\pi3$ and $\frac{dx}{d\theta}=-2\sqrt3$ , so $\frac{d\theta}{dt}=\frac{15}{-2\sqrt3}=-\frac{5\sqrt3}2$ .

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

Upper half of the ellipse
Tangent at Q
Normal at Q
A right angle between the tangent and normal when the guide is enabled

Key idea

Connected rates multiply: $\frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot\frac{dx}{dt}$ .

Ideas to try

Slide Q to the left end. What is

\theta

there?

(b)

Tangent to a parametric curve

$P$ has parameter $p$ with $0<p<\frac\pi2$ . Show that the tangent at $P$ is $y=-\frac34\cot p\,x+3\operatorname{cosec}p$ .

Hint

$\frac{dy}{dx}=\frac{3\cos\theta}{-4\sin\theta}$ at $\theta=p$ .
Use the point–gradient form and simplify with $\sin^2p+\cos^2p=1$ .

Answer · (b)

y=-(3/4)x\cot p+3\csc p

(c)

Intercepts and area

The tangent at $P$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the area of triangle $OAB$ .

Explore this idea

Hint

$B$ : put $x=0$ . $A$ : put $y=0$ .
Area $=\frac12\cdot OA\cdot OB$ .

Answer · (c)

=12/\sin2p

Explore the idea

The tangent cuts off a triangle

Move P. The tangent meets the axes at A and B; the triangle OAB has area 12/sin 2p.

Position along the curve(2.82843, 2.12132)p = 0.7854

Try an example

Tangent gradient

-0.75

At $p=\frac\pi4$ the area is $\frac{12}{\sin(\pi/2)}=12$ , the smallest possible: $\sin2p\le1$ .

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

Quarter-ellipse
Tangent at P

Key idea

The area $\frac{12}{\sin2p}$ is smallest (12) when $p=\frac\pi4$ .

Ideas to try

At which

p

is the area smallest?

(d)

Locus by eliminating a parameter

Let $M$ be the midpoint of $AB$ . Find a cartesian equation of the curve traced by $M$ as $p$ varies.

Explore this idea

Hint

$M=\left(\frac2{\cos p},\frac3{2\sin p}\right)$ .
Eliminate $p$ with $\sin^2p+\cos^2p=1$ .

Answer · (d)

4/x^2+9/(4y^2)=1

Explore the idea

The midpoint traces a curve with asymptotes x = 2 and y = 3/2

Switch on the asymptotes and some midpoints.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

Locus of M
Its asymptotes

Ideas to try

Why does the locus stay above

y=\frac32

and to the right of

x=2

Ellipse tangent intercepts and their midpoint

Solution

Connected rates of change

The parameter

Derivative of $x$

Chain rule

Evaluate

Q moves along the ellipse

Tangent to a parametric curve

Gradient

Tangent through $P(4\cos p,\,3\sin p)$

Expand

Combine

Intercepts and area

$B$

$A$

Area

The tangent cuts off a triangle

Locus by eliminating a parameter

Midpoint

Express the trig functions

Eliminate

Range of the locus

The midpoint traces a curve with asymptotes x = 2 and y = 3/2

The parameter

Derivative of xxx

Chain rule

Evaluate

Gradient

Tangent through P(4cos⁡p, 3sin⁡p)P(4\cos p,\,3\sin p)P(4cosp,3sinp)

Expand

Combine

BBB

AAA

Area

Midpoint

Express the trig functions

Eliminate

Range of the locus

Derivative of $x$

Tangent through $P(4\cos p,\,3\sin p)$

$B$

$A$