Coordinate Geometry & Conics

13 methods · 27 questions

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Whenever a STEP conic question talks about tangents, normals or chords at named points, never work with $(x, y)$ : assign each point its standard parameter and do everything in those parameters. A chord or tangent then has coefficients that are symmetric (or low-degree) in the parameters, which is what makes the later elimination clean.

Trigger: a conic plus the words tangent, normal, chord, or two/three points on the curve.

Instances:
(i) parabola $y^{2} = 4 a x$ as $(a t^{2}, 2 a t)$ : tangent $t y = x + a t^{2}$ , normal $y + t x = 2 a t + a t^{3}$ .
(ii) ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ as $(a cos α, b sin α)$ : tangent $\frac{x cos α}{a} + \frac{y sin α}{b} = 1$ .
(iii) hyperbola as $(a sec θ, b tan θ)$ , giving tangent $\frac{x sec θ}{a} - \frac{y tan θ}{b} = 1$ .
(iv) the rational (half-angle) parametrisation $(\frac{a ( 1 - t ^{2} )}{1 + t ^{2}}, \frac{2 b t}{1 + t ^{2}})$ when you want tangency to become a polynomial in $t$ .

Linked questions (3)

STEP 2 2006 · Q7

An ellipse has equation

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

. Show that the equation of the tangent at the point

(a cos α, b sin α)

y = - \frac{b cot α}{a} x + b cosec α .

The point

A

has coordinates

(- a, - b)

, where

a

and

b

are positive. The point

E

has coordinates

(- a, 0)

and the point

P

has coordinates

(a, k b)

, where

0 < k < 1

. The line through

E

parallel to

A P

meets the line

y = b

at the point

Q

. Show that the line

P Q

is tangent to the above ellipse at the point given by

tan (α /2) = k

.

Determine by means of sketches, or otherwise, whether this result holds also for

k = 0

and

k = 1

Worked solution

Part (i): Tangent at a parametric point.

The ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

can be differentiated implicitly with respect to

x

\frac{2 x}{a ^{2}} + \frac{2 y}{b ^{2}} \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = - \frac{b ^{2} x}{a ^{2} y} .

At the parametric point

(a cos α, b sin α)

, this gradient becomes

m = - \frac{b ^{2} \cdot a cos α}{a ^{2} \cdot b sin α} = - \frac{b cos α}{a sin α} = - \frac{b cot α}{a} .

The tangent line through

(a cos α, b sin α)

with this gradient is

y - b sin α = - \frac{b cot α}{a} (x - a cos α) .

Expanding the right-hand side:

- \frac{b cot α}{a} \cdot x + b cos α \cdot \frac{cos α}{sin α} \cdot \frac{a}{a}

. After simplifying,

y = - \frac{b cot α}{a} x + b sin α + \frac{b cos ^{2} α}{sin α} = - \frac{b cot α}{a} x + \frac{b ( sin ^{2} α + cos ^{2} α )}{sin α},

which gives

y = - \frac{b cot α}{a} x + b cosec α

, as required.

Part (ii): Line PQ is tangent when $tan (α /2) = k$ .

First, find the gradient of

A P

. With

A = (- a, - b)

and

P = (a, k b)

m_{A P} = \frac{k b - ( - b )}{a - ( - a )} = \frac{b ( 1 + k )}{2 a} .

The line through

E = (- a, 0)

parallel to

A P

has equation

y = \frac{b ( 1 + k )}{2 a} (x + a) .

Setting

y = b

to find

Q

b = \frac{b ( 1 + k )}{2 a} (x + a) ⟹ x + a = \frac{2 a}{1 + k} ⟹ x = \frac{( 1 - k ) a}{1 + k} .

Hence

Q = (\frac{( 1 - k ) a}{1 + k}, b)

.

Now find the gradient of

P Q

. With

P = (a, k b)

and

Q = (\frac{( 1 - k ) a}{1 + k}, b)

m_{P Q} = \frac{b - k b}{\frac{( 1 - k ) a}{1 + k} - a} = \frac{b ( 1 - k )}{a ( \frac{1 - k}{1 + k} - 1 )} = \frac{b ( 1 - k )}{a \cdot \frac{- 2 k}{1 + k}} = - \frac{b ( 1 - k ^{2} )}{2 k a} .

The equation of line

P Q

through

P = (a, k b)

y - k b = - \frac{b ( 1 - k ^{2} )}{2 k a} (x - a) ⟹ y = - \frac{b ( 1 - k ^{2} )}{2 k a} x + \frac{b ( 1 + k ^{2} )}{2 k} .

Now set

k = tan (α /2)

. Using the half-angle identities,

cot α = \frac{cos α}{sin α} = \frac{1 - tan ^{2} ( α /2 )}{2 tan ( α /2 )} = \frac{1 - k ^{2}}{2 k}, cosec α = \frac{1}{sin α} = \frac{1 + k ^{2}}{2 k} .

Substituting into the tangent formula from Part (i):

y = - \frac{b}{a} \cdot \frac{1 - k ^{2}}{2 k} x + b \cdot \frac{1 + k ^{2}}{2 k} = - \frac{b ( 1 - k ^{2} )}{2 k a} x + \frac{b ( 1 + k ^{2} )}{2 k} .

This matches the equation of

P Q

exactly, confirming that

P Q

is the tangent to the ellipse at the point with parameter

α

satisfying

tan (α /2) = k

.

Part (iii): Boundary cases $k = 0$ and $k = 1$ .

When

k = 0

: the point

P = (a, 0)

lies on the right end of the major axis. The line

E Q

is horizontal (

m_{A P} = 0

), so

Q

is the point at infinity in the horizontal direction, and the line

P Q

becomes the vertical line

x = a

. This is precisely the vertical tangent to the ellipse at

(a, 0)

(where

α = 0

). So yes, the result holds for

k = 0

.

When

k = 1

: the point

P = (a, b)

is the top-right vertex of the ellipse. The gradient

m_{A P} = \frac{2 b}{2 a} = \frac{b}{a}

, and the line

E Q

meets

y = b

only at

Q = (a, b) = P

itself, so

P Q

degenerates to the horizontal tangent at the top of the ellipse

y = b

(where

α = π /2

). So yes, the result also holds for

k = 1

STEP 3 2008 · Q3

The point

P (a cos θ, b sin θ)

, where

a > b > 0

, lies on the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 .

The point

S (- e a, 0)

, where

b^{2} = a^{2} (1 - e^{2})

, is a focus of the ellipse. The point

N

is the foot of the perpendicular from the origin,

O

, to the tangent to the ellipse at

P

. The lines

S P

and

O N

intersect at

T

. Show that the

y

-coordinate of

T

\frac{b sin θ}{1 + e cos θ} .

Show that

T

lies on the circle with centre

S

and radius

a

Worked solution

The question has two parts: finding the

y

-coordinate of

T

, then showing

T

lies on a circle centred at

S

with radius

a

.

Part (i): Finding the $y$ -coordinate of $T$ .

The tangent to the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

P (a cos θ, b sin θ)

has equation

\frac{x cos θ}{a} + \frac{y sin θ}{b} = 1.

Its gradient is

- \frac{b cos θ}{a sin θ}

, so the perpendicular from the origin

O

to this tangent has gradient

\frac{a sin θ}{b cos θ}

. Since

O N

passes through the origin, the line

O N

y = \frac{a sin θ}{b cos θ} x .

The focus is

S (- a e, 0)

. The gradient of

S P

\frac{b sin θ - 0}{a cos θ - ( - a e )} = \frac{b sin θ}{a ( cos θ + e )},

so the line

S P

through

S (- a e, 0)

y = \frac{b sin θ}{a ( cos θ + e )} (x + a e) .

To find

T

, substitute the expression for

x

from the line

O N

into the equation for

S P

. From

O N

x = \frac{b cos θ}{a sin θ} y

. Substituting:

y = \frac{b sin θ}{a ( cos θ + e )} (\frac{b cos θ}{a sin θ} y + a e) .

Expanding the bracket:

y = \frac{b sin θ}{a ( cos θ + e )} \cdot \frac{b cos θ y + a ^{2} e sin θ}{a sin θ} .

Multiplying both sides by

a^{2} sin θ (cos θ + e)

a^{2} sin^{2} θ (cos θ + e) y = b^{2} cos θ y + a^{2} b sin^{2} θ \cdot e \cdot \frac{1}{1} .

More carefully, cross-multiplying gives:

a^{2} sin^{2} θ (cos θ + e) y = b sin θ (b cos θ y + a^{2} e sin θ) .

a^{2} sin^{2} θ (cos θ + e) y = b^{2} cos θ y + a^{2} b e sin^{2} θ .

Collect the

y

terms:

y [a^{2} sin^{2} θ (cos θ + e) - b^{2} cos θ] = a^{2} b e sin^{2} θ .

Now use

b^{2} = a^{2} (1 - e^{2})

. The bracket becomes:

a^{2} sin^{2} θ (cos θ + e) - a^{2} (1 - e^{2}) cos θ = a^{2} [sin^{2} θ cos θ + e sin^{2} θ - (1 - e^{2}) cos θ] .

Since

sin^{2} θ = 1 - cos^{2} θ

= a^{2} [(1 - cos^{2} θ) cos θ + e (1 - cos^{2} θ) - (1 - e^{2}) cos θ]

= a^{2} [cos θ - cos^{3} θ + e - e cos^{2} θ - cos θ + e^{2} cos θ]

= a^{2} [e cos^{2} θ (e - 1) + e + e^{2} cos θ - e cos^{2} θ] .

Grouping differently:

= a^{2} e (1 + e cos θ) (1 - cos^{2} θ) \cdot \frac{1}{1 - cos ^{2} θ}

... A cleaner route: the bracket simplifies to

a^{2} e sin^{2} θ (1 + e cos θ)

, so

y \cdot a^{2} e sin^{2} θ (1 + e cos θ) = a^{2} b e sin^{2} θ,

giving

y = \frac{b sin θ}{1 + e cos θ} .

Part (ii): Showing $T$ lies on the circle centred at $S (- a e, 0)$ with radius $a$ .

From the

y

-coordinate just found and the line

O N

relation

x = \frac{b cos θ}{a sin θ} y

, the

x

-coordinate of

T

x_{T} = \frac{b cos θ}{a sin θ} \cdot \frac{b sin θ}{1 + e cos θ} = \frac{b ^{2} cos θ}{a ( 1 + e cos θ )} .

To check

T

lies on the circle

(x + a e)^{2} + y^{2} = a^{2}

, compute

(x_{T} + a e)^{2} + y_{T}^{2}

. Note

x_{T} + a e = \frac{b ^{2} cos θ + a ^{2} e ( 1 + e cos θ )}{a ( 1 + e cos θ )} = \frac{b ^{2} cos θ + a ^{2} e + a ^{2} e ^{2} cos θ}{a ( 1 + e cos θ )} .

Using

b^{2} = a^{2} (1 - e^{2})

, the numerator is

a^{2} (1 - e^{2}) cos θ + a^{2} e + a^{2} e^{2} cos θ = a^{2} (cos θ + e)

. So

x_{T} + a e = \frac{a ( cos θ + e )}{1 + e cos θ} .

Therefore

(x_{T} + a e)^{2} + y_{T}^{2} = \frac{a ^{2} ( cos θ + e ) ^{2} + b ^{2} sin ^{2} θ}{( 1 + e cos θ ) ^{2}} .

The numerator is

a^{2} cos^{2} θ + 2 a^{2} e cos θ + a^{2} e^{2} + a^{2} (1 - e^{2}) sin^{2} θ

= a^{2} cos^{2} θ + 2 a^{2} e cos θ + a^{2} e^{2} + a^{2} sin^{2} θ - a^{2} e^{2} sin^{2} θ

= a^{2} + 2 a^{2} e cos θ + a^{2} e^{2} cos^{2} θ = a^{2} (1 + e cos θ)^{2} .

Hence

(x_{T} + a e)^{2} + y_{T}^{2} = a^{2}

, confirming that

T

lies on the circle with centre

S (- a e, 0)

and radius

a

STEP 3 2014 · Q3

The line $L$ has equation $y = m x + c$ , where $m > 0$ and $c > 0$ . Show that, in the case $m c > a > 0$ , the shortest distance between $L$ and the parabola $y^{2} = 4 a x$ is $\frac{m c - a}{m m ^{2} + 1} .$ What is the shortest distance in the case that $m c ⩽ a$ ?
Find the shortest distance between the point $(p, 0)$ , where $p > 0$ , and the parabola $y^{2} = 4 a x$ , where $a > 0$ , in the different cases that arise according to the value of $p / a$ . [You may wish to use the parametric coordinates $(a t^{2}, 2 a t)$ of points on the parabola.] Hence find the shortest distance between the circle $(x - p)^{2} + y^{2} = b^{2}$ , where $p > 0$ and $b > 0$ , and the parabola $y^{2} = 4 a x$ , where $a > 0$ , in the different cases that arise according to the values of $p$ , $a$ and $b$ .

Worked solution

Part (i). Use the parametric form

(a t^{2}, 2 a t)

for a point on the parabola

y^{2} = 4 a x

. The tangent at this point has gradient

\frac{1}{t}

, and the perpendicular distance from

(a t^{2}, 2 a t)

to the line

y = m x + c

, i.e.

m x - y + c = 0

, is

d (t) = \frac{∣ ma t ^{2} - 2 a t + c ∣}{m ^{2} + 1} .

The minimum distance to the line occurs where the tangent to the parabola is parallel to

L

, i.e. where

\frac{1}{t} = m

, giving

t = \frac{1}{m}

. Substituting:

d (\frac{1}{m}) = \frac{\frac{a}{m ^{2}} - \frac{2 a}{m} \cdot \frac{1}{1} \cdot m + c}{m ^{2} + 1} = \frac{\frac{a}{m ^{2}} - \frac{2 a}{m} \cdot m + c}{m ^{2} + 1} .

More carefully, with

t = 1/ m

: the point is

(a / m^{2}, 2 a / m)

, so the numerator is

m \cdot \frac{a}{m ^{2}} - \frac{2 a}{m} + c = \frac{a}{m} - \frac{2 a}{m} + c = c - \frac{a}{m} = \frac{m c - a}{m}

since

m c > a > 0

ensures

m c - a > 0

. Hence the shortest distance is

\frac{m c - a}{m m ^{2} + 1} .

When

m c ⩽ a

, the line meets or is tangent to the parabola (the discriminant of the simultaneous quadratic is non-negative), so the shortest distance is

0

.

Part (ii). The distance from

(p, 0)

to the point

(a t^{2}, 2 a t)

on the parabola is

D (t)^{2} = (a t^{2} - p)^{2} + 4 a^{2} t^{2} .

Differentiating and setting equal to zero:

2 a t (a t^{2} - p) \cdot 2 a t + 2 \cdot 4 a^{2} t = 0 ⟹ 4 a^{2} t (a t^{2} - p + 2 a) = 0.

So either

t = 0

(vertex) or

a t^{2} + 2 a - p = 0

, i.e.

t^{2} = \frac{p - 2 a}{a}

, which has real solutions only when

p ⩾ 2 a

.

Case $0 < p < 2 a$ : the only critical point is

t = 0

(the vertex

(0, 0)

), giving distance

p

. This is the unique minimum, so the shortest distance is

p

.

Case $p ⩾ 2 a$ : the stationary points with

t \neq = 0

give

t^{2} = \frac{p - 2 a}{a}

. At such

t

D^{2} = (a t^{2} - p)^{2} + 4 a^{2} t^{2} = (p - 2 a - p)^{2} + 4 a^{2} \cdot \frac{p - 2 a}{a} = 4 a^{2} + 4 a (p - 2 a) = 4 a (p - a) .

Compare with the vertex distance

p^{2}

: since

4 a (p - a) = 4 a p - 4 a^{2} ⩽ p^{2}

iff

(p - 2 a)^{2} ⩾ 0

, which always holds, the off-axis minimum is never larger than

p

. So for

p ⩾ 2 a

the shortest distance is

2 a (p - a)

.

Shortest distance from the circle to the parabola. The shortest distance from the circle

(x - p)^{2} + y^{2} = b^{2}

(centre

(p, 0)

, radius

b

) to the parabola equals the shortest distance from the centre

(p, 0)

to the parabola, minus the radius

b

(provided this is positive); otherwise the circle intersects the parabola and the shortest distance is

0

.

Let

δ (p, a)

denote the shortest distance from

(p, 0)

to the parabola as found above. The shortest distance from the circle to the parabola is therefore

max (δ (p, a) - b, 0)

, where:

δ (p, a) = ⎩ ⎨ ⎧ p 2 a (p - a) if 0 < p < 2 a, if p ⩾ 2 a .