Ellipse: Chord, Tangent and Normal

EQUATION OF THE TANGENT AT A POINT OF AN ELLIPSE

(i) Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Slope of tangent to the ellipse at a point (x₁, y₁) = ${\left[\frac{dy}{dx}\right]}_{\left(x_1,y_1\right)}=-\frac{b^2}{a^2}\frac{{\text{x}}_{\text{1}}^{}}{{\text{y}}_{\text{1}}^{}}$

Hence the equation of the tangent at (x₁, y₁) is y -y₁ =$\frac{-b^2{x}_1}{a^2{y}_1}\left(x-x_1\right)$

$\Rightarrow$ $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$ i.e. T= 0

(ii) Equation of tangent at the point $\theta$ i.e. (a cos$\theta$, b sin$\theta$) is obtained by putting

x₁ = a cos$\theta$, y₁ = b sin$\theta$; $\Rightarrow$ $\frac{x\text{ cos}\theta }{a}+\frac{y\text{ sin}\theta }{b}=1$

EQUATION OF THE TANGENT IN TERMS OF ITS SLOPE; USING THE CONCEPT OF COMPARISON:

The equations of the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}$=1 with slope m are

y = mx $\pm$ $\sqrt{a^2{m}^2+b^2}$ for all finite values of m.

Moreover the line touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1at\left(\frac{\pm a^{2}m}{\sqrt{\left(a^2{m}^2+b^2\right)}},\frac{\pm b^2}{\sqrt{\left(a^2{m}^2+b^2\right)}}\right)$.

Illustration 1: From a point P two tangents are drawn one each to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1and\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$, If tangents are perpendicular to each other, then find locus of P.

Solution: The tangent at (a cos$\theta$, b sin$\theta$) on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ ….(i)

The tangent at (a cos$\varphi$, b sin$\varphi$) on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$\frac{x}{a}\cos \varphi +\frac{y}{b}\sin \varphi =1$ ….(ii)

(i) and (ii) are perpendicular cos ($\theta$ - $\varphi$) = 0 or $\theta$ = $\varphi$ + $\frac{\pi }{2}$

$-\frac{x}{a}\sin \varphi +\frac{y}{b}\cos \varphi =1$ ….(iii)

Eliminating $\varphi$ from (ii) and (iii)

Locus is (x² + y²)² = b² (x + y)² + a²(x - y)²

EQUATION OF THE PAIR OF TANGENTS DRAWN FROM THE POINT (X₁, Y₁)

The following equation will be the equation of the pair of tangents from (x₁, y₁).

${\left\{\frac{x_1\left(x-x_1\right)}{a^2}+\frac{y_1\left(y-y_1\right)}{b_2}\right\}}^2=\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\right)\left\{\frac{{\left(x-x_1\right)}^2}{a^2}+\frac{{\left(y-y_1\right)}^2}{b^2}\right\}$

${\left\{\left(\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b_2}\right)-\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)\right\}}^2=\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\right)$ $\left\{\frac{x_{}^2}{a^2}+\frac{y_{}^2}{b^2}-2\left(\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b_2}\right)+\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right\}$

If we assume that T = $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b_2}$ – 1, S = $\frac{x_{}^2}{a^2}+\frac{y_{}^2}{b^2}-1$, and

S₁ = $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$, then equation becomes

(T – S₁)² = S₁(S – 2T + S₁)

i.e. T² = SS₁.

Illustration 2: Find the angle between the pair of tangents drawn to the ellipse 3x² + 2y² = 5 from the point (1, 2).

Solution: Let the equation of the line passing through (1, 2) be

y – 2 = m (x – 1) or y = mx – m + 2 … (1)

Line (1) touches 3x² + 2y² = 5 if

3x² + 2(mx – m + 2)² = 5

or (3 + 2m²)x² – 4m(m–2)x + 2m² + 8 – 8m – 5 = 0

or (3 + 2m²)x² – 4m(m–2)x + 2m² – 8m +3 = 0

For equal roots. D = 0

$\Rightarrow$ [4m(m–2)]² – 4(3 + 2m²)(2m² – 8m + 3) = 0

or 4m² (m² – 4m + 4) – (6m² – 24m + 9 + 4m⁴ – 16m³ + 6m²) = 0

or 4m⁴ – 16m³ + 16m² –12m² – 4m⁴ + 16m³ + 24m – 9 = 0

or 4m² + 24m – 9 = 0

$\Rightarrow$ m₁ + m₂ = –6 and m₁m₂ = –9/4

tan$\theta$ = $\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{36+9}}{1+9/4}\right|=\left|\frac{3\sqrt{5}}{-5}\times 4\right|=\left|\frac{12}{\sqrt{5}}\right|$

$\Rightarrow$ $\theta$ = tan^–1$\left|\frac{12}{\sqrt{5}}\right|$

Alternative Method 1:

Hint: Find the equation of pair of tangent i.e. T² = SS₁,

Find the angle between above two tangents i.e. tan$\theta$ = $\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Alternative Method 2:

Equation to the ellipse is $\frac{x^2}{5/3}+\frac{y^2}{5/2}=1$

Let the equation to tangent be y = mx + $\sqrt{\frac{5}{3}{m}^2+\frac{5}{2}}$

Which is passing through (1, 2) $\Rightarrow$ 2 = m + $\sqrt{\frac{5}{3}{m}^2+\frac{5}{2}}$

$\Rightarrow$ (2 – m)² = $\frac{5}{3}{m}^2+\frac{5}{2}$ $\Rightarrow$ 4m² + 24m – 9 = 0 has two roots m₁ and m₂ which are the slopes of the tangents from the point (1, 2)

$\Rightarrow$ m₁ + m₂ = –6 ; m₁m₂ = –9/4

$\therefore$ Angle between the tangents is tan^–1$\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|={\tan }^{-1}\left|\frac{\sqrt{45}}{1-9/4}\right|={\tan }^{-1}\left|\frac{12}{\sqrt{5}}\right|$

Director Circle

The locus of the point of intersection of a pair of perpendicular tangents to an ellipse is called Director Circle.

EQUATION OF DIRECTOR CIRCLE

x² + y² = a² + b²

EQUATION OF THE NORMAL AT A POINT OF AN ELLIPSE

Equation of the normal is y-y₁ = $\frac{a^2}{b^2}\frac{{\text{y}}_{\text{1}}}{{\text{x}}_{\text{1}}}\left(x-x_1\right)$ $\Rightarrow$ $\frac{x-x_1}{x_1/a^2}=\frac{y-y_1}{y_1/b^2}$

$\bullet$ Equation of normal at (a cos$\theta$, b sin$\theta$) is $\frac{x-a\cos \theta }{\frac{a\cos \theta }{a^2}}=\frac{y-b\sin \theta }{\frac{b\sin \theta }{b^2}}=\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }$ = a² – b²

$\Rightarrow$ ax sec$\theta$ - by cosec$\theta$ = a² - b²

Example -3: If the normals to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) are concurrent, prove that

$\left|\begin{array}{c}{x}_1{y}_1{x}_1{y}_1\\ x_2{y}_2{x}_2{y}_2\\ x_3{y}_3{x}_3{y}_3\end{array}\right|$

= 0.

Solution: The equation of the normal to the given ellipse at (x₁, y₁) is

a²xy₁ – b²yx₁ – (a² – b²)x₁y₁ = 0 . . .(1)

Similarly the normals at (x₂, y₂) and (x₃, y₃) are

a²xy₂ – b²yx₂ – (a² – b²)x₂y₂ = 0 …(2)

a²xy₃ – b²yx₃ – (a² – b²)x₃y₃ = 0 …(3)

Eliminating a²x, b²y and (a² - b²) from (1), (2) and (3), we find that the three lines are concurrent if

$\Rightarrow$ $\left|\begin{array}{c}{y}_1{x}_1{x}_1{y}_1\\ y_2{x}_2{x}_2{y}_2\\ y_3{x}_3{x}_3{y}_3\end{array}\right|$

= 0 $\Rightarrow$ $\left|\begin{array}{c}{x}_1{y}_1{x}_1{y}_1\\ x_2{y}_2{x}_2{y}_2\\ x_3{y}_3{x}_3{y}_3\end{array}\right|$ = 0

EQUATION OF MID POINT CHORD

Equation of mid chord whose mid point is (x₁,y₁) is given by

T=S₁ where T=$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}-1$ and S₁=$\frac{{x_1}^2}{a^2}+\frac{{y_1}^2}{b^2}-1$.

Illustration 4: Find the locus of the midpoint of chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, that are parallel to the line y = 2x + c.

Solution: Let the mid – point of the chord be (h, k). Then equation of this chord will be

T = S₁

$\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{k^2}+\frac{k^2}{b^2}$

Slope of line $\frac{-h/a^2}{k/b^2}=2$

Required locus is $y=-\frac{b^2}{2a^2}x$

CHORD OF CONTACT

Equation of chord of contact is given by T = 0, where T = $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}-1$.

Illustration 5: From a point O on the circle x² + y² = 25, tangents OP and OQ are drawn to the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ Show that the locus of the mid point of the chord PQ describes the curve x² + y² = 25${\left[\frac{x^2}{4}+\frac{y^2}{1}\right]}^2$.

Solution: O $\equiv$ (5 cosq, 5 sin$\theta$)

Let R be middle point of PQ. Let R $\equiv$ (h, k) .

Equation of the chord of contact PQ is T = 0

i.e. $\frac{5x\cos \theta }{4}+\frac{5y\sin \theta }{1}=1$ . . . . . (1)

Equation of the chord PQ with middle point R(h, k) is

T = S₁ i.e. $\frac{xh}{4}+\frac{yk}{1}-1=\frac{h^2}{4}+\frac{k^2}{1}-1$

or $\frac{xh}{4}+\frac{yk}{1}=\frac{h^2}{4}+\frac{k^2}{1}$ . . . .(2)

(1) and (2) are the same equations.

$\Rightarrow$ $\frac{5\cos \theta }{h}=\frac{5\sin \theta }{k}=\frac{1}{\frac{h^2}{4}+\frac{k^2}{1}}$ $\Rightarrow$ $\frac{\left(h^2+k^2\right)}{{\left(\frac{h^2}{4}+\frac{k^2}{1}\right)}^2}.\frac{1}{25}$ = 1

Hence locus of (h, k) is (x² +y²) = 25${\left[\frac{x^2}{4}+\frac{y^2}{1}\right]}^2$.