In an ellipse, x2/a2 + y2/b2 = 1, a b, if the focal distances of a point on the ellipse - Maths - Conic Sections

Question

In an ellipse, x2/a2 +
y2/b2 = 1, a>b, if the focal distances of
a point on the ellipse ( the point is neither on X-axis nor on
Y-axis ) are r1and r2, then find the length
of the normal drawn at this point ( length of normal means the
length of the portion of the normal between the point and the major
axis ) ?

Lalit Mehra · Accepted Answer

The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$

∴ x- axis is the major axis of the ellipse.

Let S(ae, 0) and S' (– ae, o) be the foci of the ellipse, where e is the eccentricity of the ellipse.

Let P(a cos θ, b sin θ) be point on the ellipse.

Given, r₁ = a(1 – e cos θ) and r₂ = a(1 + e cos θ)

Equation of normal at P(a cos θ, b sin θ) is ax sec θ – by cosec θ = a²– b²...(1)

On the x- axis, y = 0

∴ ax sec θ – 0 = a² – b²

$\Rightarrow x = \frac{a^{2} - b^{2}}{a} \cos θ$

∴ Normal intersect the major axis at $(\frac{a^{2} - b^{2}}{a} \cos θ , 0)$

Length of normal

$= \sqrt{(\frac{a^{2} - b^{2}}{a} \cos θ - a \cos θ)^{2} + (0 - b \sin θ)^{2}} = \sqrt{(\frac{a^{2} \cos θ - b^{2} \cos θ - a^{2} \cos θ}{a})^{2} + (- b \sin θ)^{2}} = \sqrt{\frac{b^{4} \cos^{2} θ}{a^{2}} + b^{2} \sin^{2} θ} = \frac{b}{a} \sqrt{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{b}{a} \sqrt{a^{2} (1 - e^{2}) \cos^{2} θ + a^{2} \sin^{2} θ} (∵ b^{2} = a^{2} (1 - e^{2})^{}) = \frac{b}{a} \sqrt{a^{2} \cos^{2} θ - a^{2} e^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{b}{a} \sqrt{a^{2} - a^{2} e^{2} \cos^{2} θ} = \frac{b}{a} \sqrt{a^{2} (1 - e^{2} \cos^{2} θ)} = b \sqrt{1 - e^{2} \cos^{2} θ} = b \sqrt{(1 - e \cos θ) (1 + e \cos θ)} = b \sqrt{(\frac{r_{1}}{a}) (\frac{r_{2}}{a})} = \frac{b}{a} \sqrt{r_{1} r_{2}}$

Thus, length of the normal is $\frac{b}{a} \sqrt{r_{1} r_{2}}$ .

9