# Equation of the Chord – Ellipse

## Equation of the Chord – Ellipse

Equation of the chord of the ellipse whose midpoint is (x₁, y₁).

Let the slope of the chord be tan. Then any point on the chord at distance r from the point (x₁, y₁) is (x + rcosθ, y + rsinθ).

If this point lies on the ellipse, then $$\frac{{{({{x}_{1}}+r\cos \theta )}^{2}}}{{{a}^{2}}}+\frac{{{({{y}_{1}}+r\cos \theta )}^{2}}}{{{b}^{2}}}=1$$.

Since the line cuts the ellipse at two-point Q and R, this is quadratic in r, whose roots are r₁ = PQ and r₂ = – QR.

Hence the sum of roots.

r₁ + r₂ = 0

(As PQ = QR)

Then, from (i), the co-efficient of r is 0.

∴ $$\frac{2{{x}_{1}}\cos \theta }{{{a}^{2}}}+\frac{2{{y}_{1}}\sin \theta }{{{b}^{2}}}=0$$.

$$\tan \theta =-\frac{{{b}^{2}}{{x}_{1}}}{{{a}^{2}}{{y}_{1}}}$$.

Hence, the equation of chord is $$y-{{y}_{1}}=-\frac{{{b}^{2}}{{x}_{1}}}{{{a}^{2}}{{y}_{1}}}(x-{{x}_{1}})$$.

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

T = S₁

Where, $${{S}_{1}}=\frac{{{x}_{1}}^{2}}{{{a}^{2}}}+\frac{{{y}_{1}}^{2}}{{{b}^{2}}}-1$$.

Example: Find the length of the chord of the ellipse $$\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1$$ whose middle point is (½, ⅖).

Solution: The equation of the chord having (½, ⅖) as mid-point $$\frac{1/4}{25}+\frac{4/25}{16}-1=\frac{(1/2)x}{25}+\frac{(2/5)y}{16}-1$$ (∵ $$\frac{x{{x}_{1}}}{{{a}^{2}}}+\frac{y{{y}_{1}}}{{{b}^{2}}}-1=\frac{{{x}_{1}}^{2}}{{{a}^{2}}}+\frac{{{y}_{1}}^{2}}{{{b}^{2}}}-1$$ (or) T = S₁).

4x + 5y = 4

5y = 4(1 – x)

Solving with ellipse, we get

16x² + 16 (1 – x)² = 400

x² – x – 12 = 0

x = 4, -3

x = 4, y = – 12/3

x = – 3, y = 16/5

∴ The length of the chord is $$\sqrt{\{{{7}^{2}}+{{(\frac{28}{5})}^{2}}\}}=7\sqrt{\frac{41}{25}}$$.