Let d be the perpendicular distance from the centre of the ellipse x²/a² + y²/b² = 1 to the tangent

Q: Why do we use $a\cos\theta$ and $b\sin\theta$ instead of just $x$ and $y$?

Great question! Using Cartesian coordinates $x$ and $y$ forces you to carry the constraint $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ throughout the entire calculation. Parametric coordinates automatically satisfy this constraint, turning a messy algebraic problem into a clean trigonometric one.

Q: I got stuck on the focal distance part. Is there a simpler way to remember $PF_1 = a + ex_P$?

Think of it as the definition of the ellipse. The distance from the focus is essentially the distance from the directrix scaled by eccentricity. It's a fundamental property derived from the focus-directrix definition. Memorizing it saves precious minutes in the exam hall.

Q: Why does the $b^2/d^2$ term simplify so nicely? I keep getting lost in the algebra.

The key is to distribute the $b^2$ carefully. When you multiply $b^2$ by $\frac{\sin^2\theta}{b^2}$, the $b^2$ cancels out. That cancellation is the 'aha!' moment. If you don't see it, pause, write it out step-by-step, and look for the identity $\sin^2\theta + \cos^2\theta = 1$.

Question

Let $d$ be the perpendicular distance from the centre of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to the tangent drawn at a point $P$ on the ellipse. If $F_1$ and $F_2$ are the two foci of the ellipse, then show that $(PF_1 - PF_2)^2 = 4a^2 \left( 1 - \frac{b^2}{d^2} \right)$.

Sigma Percentile · Accepted Answer

### The Elegance of the Ellipse: A Geometric Journey

Imagine you are standing at the center of an ellipse, looking out at a point $P$ dancing along its perimeter. This isn't just a curve; it is a perfect balance of geometry and algebra.

Today, we are going to unravel a beautiful relationship involving the tangent at $P$ and the two foci, $F_1$ and $F_2$. Let's take a breath and walk through this derivation together.

### Phase 1: Setting the Stage

We begin with the standard ellipse equation:

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

The foci, $F_1(-ae, 0)$ and $F_2(ae, 0)$, are the heart of this shape. To make our journey easier, we choose a point $P$ on the ellipse using parametric coordinates: $P(a\cos	heta, b\sin	heta)$.

These coordinates automatically satisfy the ellipse equation, saving us from algebraic nightmares later. Using the point-of-contact form, the equation of the tangent line at $P$ is:

$$ \frac{x}{a}\cos	heta + \frac{y}{b}\sin	heta = 1 $$

### Phase 2: The Perpendicular Distance

The problem introduces $d$, the perpendicular distance from the center $(0,0)$ to this tangent. Using the standard formula for the distance from the origin to a line $Ax + By + C = 0$, which is $\frac{|C|}{\sqrt{A^2 + B^2}}$, we find:

$$ d = \frac{|-1|}{\sqrt{\frac{\cos^2	heta}{a^2} + \frac{\sin^2	heta}{b^2}}} $$

Squaring and inverting this gives us a powerful tool:

$$ \frac{1}{d^2} = \frac{\cos^2	heta}{a^2} + \frac{\sin^2	heta}{b^2} $$

### Phase 3: The Focal Distances

Now, let's look at the focal distances, $PF_1$ and $PF_2$. A fundamental property of the ellipse tells us that $PF_1 = a + ex_P$ and $PF_2 = a - ex_P$.

Since our $x_P = a\cos	heta$, these become $PF_1 = a(1 + e\cos	heta)$ and $PF_2 = a(1 - e\cos	heta)$. When we subtract these, the $a$ terms vanish:

$$ PF_1 - PF_2 = 2ae\cos	heta $$

Squaring this gives us our Left Hand Side (LHS):

$$ (PF_1 - PF_2)^2 = 4a^2e^2\cos^2	heta $$

### Phase 4: The Grand Unification

Now, let's tackle the Right Hand Side (RHS): $4a^2(1 - \frac{b^2}{d^2})$. We first evaluate $\frac{b^2}{d^2}$ using our earlier expression:

$$ \frac{b^2}{d^2} = b^2 \left( \frac{\cos^2	heta}{a^2} + \frac{\sin^2	heta}{b^2} ight) = \frac{b^2}{a^2}\cos^2	heta + \sin^2	heta $$

Substitute this into the RHS expression:

$$ 1 - \left( \frac{b^2}{a^2}\cos^2	heta + \sin^2	heta ight) = (1 - \sin^2	heta) - \frac{b^2}{a^2}\cos^2	heta $$

Using the identity $1 - \sin^2	heta = \cos^2	heta$, this simplifies to $\cos^2	heta(1 - \frac{b^2}{a^2})$. Since $e^2 = 1 - \frac{b^2}{a^2}$, the expression becomes $e^2\cos^2	heta$.

Multiplying by $4a^2$, we obtain $4a^2e^2\cos^2	heta$. The LHS and RHS match perfectly, confirming the identity:

$$ \mathbf{(PF_1 - PF_2)^2 = 4a^2 \left( 1 - \frac{b^2}{d^2} ight)} $$

Visualized Solution

The Ellipse and its Foci

Point $P$ and the Tangent

Equation of the Tangent

The Perpendicular Distance $d$

Calculating $\frac{1}{d ^{2}}$

Focal Distances $P F_{1}$ and $P F_{2}$

Substituting $x_{P} = a cos θ$

Difference of Focal Distances

Evaluating the Left Hand Side (LHS)

Evaluating the Right Hand Side (RHS)

Simplifying $\frac{b ^{2}}{d ^{2}}$

Calculating $1 - \frac{b ^{2}}{d ^{2}}$

Using Trigonometric Identity

Connecting with Eccentricity

Finalizing the Proof

The Sigma Insight: Equation of Tangent and Normal

The Elegance of the Ellipse

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

The Ellipse and its Foci

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and the Tangent