A LevelFurther MathsEdexcelFurther Pure 1FlashcardsCoordinate SystemsTangent & Normals to Ellipses, Parabolas & Hyperbolas

Tangent & Normals to Ellipses, Parabolas & Hyperbolas (Edexcel A Level Further Maths: Further Pure 1): Flashcards

Exam code: 9FM0

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FrontTangents & Normals to Ellipses
The ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ has general point $(a \cos θ, b \sin θ)$ . What are the two ways of finding the gradient of the tangent there?

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The ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ has general point $(a \cos θ, b \sin θ)$ . What are the two ways of finding the gradient of the tangent there?
Implicit differentiation of the Cartesian equation, then substituting $x = a \cos θ$ and $y = b \sin θ$ into the result.
Or parametric differentiation of the two parametric equations, which reaches the same gradient without touching the Cartesian form at all.
You have the gradient $m_{T}$ of the tangent to an ellipse at $(a \cos θ, b \sin θ)$ . What do you do with it next?
Substitute it, along with $x_{1} = a \cos θ$ and $y_{1} = b \sin θ$ , into the straight-line equation $y - y_{1} = m_{T} (x - x_{1})$ .
Then simplify, which for an ellipse means clearing the fractions and using $\cos^{2} θ + \sin^{2} θ \equiv 1$ to collapse the constant term.
Fill in the condition for the line $y = m x + c$ to be a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ :
$a^{2}______+ b^{2} =______for such a line$
The completed condition is:
$a^{2} m^{2} + b^{2} = c^{2} for such a line$
The left-hand side is always positive, so every gradient $m$ gives two tangents, one with $c$ positive and one with $c$ negative, on opposite sides of the ellipse.
How do you prove that a line $y = m x + c$ is a tangent to an ellipse only when a particular condition on $m$ and $c$ holds?
Substitute the line into the ellipse's equation and rearrange into a quadratic in $x$ , whose roots are the $x$ -coordinates of the points of intersection.
A tangent touches only once, so force the discriminant to be zero, and simplifying what is left gives the condition.
True or False?
The method of substituting a gradient into $y - y_{1} = m (x - x_{1})$ works at every point of an ellipse.
False.
At the two ends of the major axis, $(\pm a, 0)$ , the tangent is vertical, and that straight-line form cannot represent a vertical line at all.
Write those two tangents down directly as $x = a$ and $x = - a$ instead of trying to compute a gradient for them.
How does finding the normal to an ellipse differ from finding the tangent?
In exactly one place: having found the tangent's gradient you take its negative reciprocal, and use that in $y - y_{1} = m_{N} (x - x_{1})$ instead.
Every other step, including the point you substitute and the identity used to tidy the result, is unchanged.
At $P (a \cos θ, b \sin θ)$ the tangent to an ellipse is $(b \cos θ) x + (a \sin θ) y = a b$ . Why is the right-hand side just $a b$ , with no $θ$ in it?
Because the two terms that land there combine as $a b (\cos^{2} θ + \sin^{2} θ)$ .
That bracket is equal to $1$ , so the whole right-hand side collapses to a constant, which is exactly the job the identity is doing.
How does implicit differentiation of $y^{2} = 4 a x$ give the gradient of the tangent at $P (a t^{2}, 2 a t)$ ?
Differentiating gives $2 y \frac{d y}{d x} = 4 a$ , so substituting $y = 2 a t$ leaves $4 a t \frac{d y}{d x} = 4 a$ .
The $a$ cancels from both sides and the gradient is simply $\frac{1}{t}$ .
Fill in the condition for the line $y = m x + c$ to be a tangent to the parabola $y^{2} = 4 a x$ :
$a =______where m is the gradient and c the intercept$
The completed condition is:
$a = m c where m is the gradient and c the intercept$
So for a given gradient $m$ there is exactly one tangent to the parabola, the line whose intercept is $c = \frac{a}{m}$ .
True or False?
The tangent to the parabola $y^{2} = 4 a x$ at its vertex is the line $y = 0$ .
False.
The tangent at the vertex $(0, 0)$ is the vertical line $x = 0$ , which is the $y$ -axis.
The line $y = 0$ is the parabola's axis of symmetry, and it cuts straight through the curve rather than touching it.
The tangent to $y^{2} = 4 a x$ at $P (a t^{2}, 2 a t)$ has gradient $\frac{1}{t}$ . What is the gradient of the normal there, and what equation does it lead to?
The normal's gradient is $- t$ , the negative reciprocal of $\frac{1}{t}$ .
Substituting that with $x_{1} = a t^{2}$ and $y_{1} = 2 a t$ and simplifying gives $t x + y = 2 a t + a t^{3}$ .
Why does simplifying the tangent to a parabola need no trigonometric or hyperbolic identity, when the other conics do?
Because the parabola's parameter is not an angle: $x = a t^{2}$ and $y = 2 a t$ are polynomials, so the algebra closes on its own.
The other conics are parametrised with $\cos$ and $\sin$ , or with $\sec$ and $\tan$ , and need an identity to collapse the terms that survive.
The tangent to $y^{2} = 4 a x$ at $P (a t^{2}, 2 a t)$ is $- x + t y = a t^{2}$ . Which point does $t = 1$ give, and what is the tangent there?
Putting $t = 1$ gives the point $(a, 2 a)$ on the parabola.
The tangent there is $- x + y = a$ , a line of gradient $1$ , which agrees with the general gradient $\frac{1}{t}$ at $t = 1$ .
A hyperbola can be parametrised with $\cosh$ and $\sinh$ , or with $\sec$ and $\tan$ . What does that mean for the equation of its tangent?
There are two equally correct forms of the same tangent: $(b \cosh θ) x - (a \sinh θ) y = a b$ and $(b \sec θ) x - (a \tan θ) y = a b$ .
They describe the same line at the same point, so which one to use is decided by the parametrisation the point was given in.
Fill in the missing sign and the missing term in the condition for $y = m x + c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ :
$a^{2} m^{2}______b^{2} =______for such a line$
The completed condition is:
$a^{2} m^{2} - b^{2} = c^{2} for such a line$
The subtraction comes straight from the minus sign in the hyperbola's own equation.
It has a consequence the other conics do not share: a real $c$ needs $a^{2} m^{2}$ to be at least $b^{2}$ , so a line whose gradient is shallower than the asymptotes can never be a tangent.
Implicit differentiation of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ gives $\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0$ . What is the tangent gradient at $(a \cosh θ, b \sinh θ)$ ?
Substituting the coordinates and rearranging gives $\frac{d y}{d x} = \frac{b \cosh θ}{a \sinh θ}$ .
It is positive for $θ > 0$ , and that sign traces back to the minus sign in the hyperbola's own equation.
True or False?
A straight line through the centre of a hyperbola can be a tangent to it.
False.
Such a line has $c = 0$ , and the tangent condition then forces $a^{2} m^{2} = b^{2}$ , so $m = \pm \frac{b}{a}$ .
Those are the gradients of the asymptotes, which approach the curve without ever touching it, so no line through the centre is a tangent.
In simplifying a hyperbola's normal, what does $\frac{\tan θ}{\sec θ}$ become?
It becomes $\sin θ$ , since $\frac{\tan θ}{\sec θ}$ is $\frac{\sin θ}{\cos θ}$ multiplied by $\cos θ$ .
That is the step that turns the normal into the tidy form $(a \sin θ) x + b y = (a^{2} + b^{2}) \tan θ$ .
At $P (a \cosh θ, b \sinh θ)$ the normal to a hyperbola is $(a \sinh θ) x + (b \cosh θ) y = (a^{2} + b^{2}) \sinh θ \cosh θ$ . Where does the $a^{2} + b^{2}$ come from?
From the two constant terms that appear once the point is substituted in and the fraction is cleared: $a^{2} \sinh θ \cosh θ$ on one side and $b^{2} \sinh θ \cosh θ$ on the other.
Moving the second across the equals sign turns its subtraction into an addition, which is where the sum of the two squares comes from.
Which extra route to the tangent gradient does $x y = c^{2}$ allow?
Its equation can be rearranged to make $y$ the subject, $y = \frac{c^{2}}{x} = c^{2} x^{- 1}$ , and then differentiated directly with the power rule.
That route works here because the equation collapses to a single simple power of $x$ , which the other conics' equations do not.
Fill in the gradient of the tangent to the rectangular hyperbola $x y = c^{2}$ at the point $(c t, \frac{c}{t})$ :
$m_{T} =______at every such point$
The completed line is:
$m_{T} = - \frac{1}{t^{2}} at every such point$
Whatever the value of $t$ , the number $t^{2}$ is positive, so every tangent to a rectangular hyperbola has a negative gradient and slopes downwards.
The tangent to $x y = c^{2}$ at $P (c t, \frac{c}{t})$ is $x + t^{2} y = 2 c t$ . Where does the $2 c t$ come from?
From substituting the point into $y - y_{1} = m_{T} (x - x_{1})$ and clearing the fraction, which leaves a term $c t$ on each side of the equals sign.
Bringing them together gives $2 c t$ , and that doubling is why the right-hand side is not simply the $x$ -coordinate of $P$ .
True or False?
Every tangent to the rectangular hyperbola $x y = c^{2}$ crosses both coordinate axes.
True.
Setting $y = 0$ in $x + t^{2} y = 2 c t$ gives $x = 2 c t$ , and setting $x = 0$ gives $y = \frac{2 c}{t}$ .
Neither can ever be zero, because $c > 0$ and $t$ is never allowed to be zero, so every tangent cuts both axes away from the origin.
The normal to $x y = c^{2}$ at $P (c t, \frac{c}{t})$ works out as $t^{3} x - t y = c (t^{4} - 1)$ . Why are the powers of $t$ so much higher than in the tangent?
Because the normal's gradient is $t^{2}$ , the negative reciprocal of the tangent's, which is already two powers up on the tangent's own gradient.
Clearing the remaining fraction $\frac{c}{t}$ then multiplies everything through by a further $t$ , and those two steps together push the equation up to $t^{3}$ and $t^{4}$ .

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