Equation Of Tangent To Ellipse

A curve in the plane which surrounds the 2 focal points such that the total of the distances to the focal point remains constant for each point on the curve. A circle is said to be a special type of an ellipse having both focal points at the same point. A line which intersects the ellipse at a point is called a tangent to the ellipse. The different forms of the tangent equation are given below:

  • Slope form of a tangent to an ellipse

If the line y = mx + c touches the ellipse x2 / a2 + y2 / b2 = 1, then c2 = a2m2 + b2. The straight line y = mx ∓ √[a2m2 + b2] represent the tangents to the ellipse.

  • Point form of a tangent to an ellipse

The equation of the tangent to an ellipse x2 / a2 + y2 / b2 = 1 at the point (x1, y1) is xx1 / a2 + yy1 / b2 = 1.

  • Parametric form of a tangent to an ellipse

The equation of the tangent at any point (a cosɸ, b sinɸ) is [x / a] cosɸ + [y / b] sinɸ.

  • Point of contact of the tangent to an ellipse

Line y = mx ∓ √[a2m2 + b2] touches the ellipse x2 / a2 + y2 / b2 = 1 at (∓a2m / √[a2m2 + b2]) , (∓b2 / √[a2m2 + b2]).

Equation Of Tangent To Ellipse Problems

Example 1: What is the locus of the point of intersection of perpendicular tangents to the ellipse x2 / a2 + y2 / b2 = 1?

Solution:

Let point be (h,k). The pair of tangents will be

(x2 / a2 + y2 / b2 − 1) (h2 / a2 + k2 / b2 − 1) = (hx / a2 +yk / b2 − 1)2

Pair of tangents will be perpendicular, if coefficient of x2 + coefficient of y2 = 0

k2 / a2b2 + h2 / a2b2 = 1 / a2 + 1 / b2

h2 + k2 = a2 + b2

Replace (h, k) by (x, y)

x2 + y2 = a2 + b2.

Example 2: What is the condition for the line lx + my − n = 0 to be tangent to the ellipse x2 / a2 + y2 / b2 = 1?

Solution:

y = [−l / m] x + n / m is tangent to x2 / a2 + y2 / b2 = 1, if

n / m = ± √b2 + [a2 / (l / m)2] or

On simplification,

n2 = m2b2 + l2a2.

Example 3: If any tangent to the ellipse x2 / a2 + y2 / b2 = 1 cuts off intercepts of length h and k on the axes, then a2 / h2 + b2 / k2 = ___________.

Solution:

The tangent at (a cosθ, b sinθ) to the ellipse is [(a cos θ) * x] / [a2] +[b sin θ] * y / b2 = 1

or x / (a / cos θ) + y / (b / sin θ) = 1

∴Intercepts are, h = a / cos θ, k = b sin θ

a2 / h2 + b2 / k2 = 1

Example 4: The equation of the tangents drawn at the ends of the major axis of the ellipse 9x2 + 5y2 − 30y = 0, are ___________.

Solution:

Change the equation 9x2 + 5y2 − 30y = 0 in standard form 9x2 + 5 (y2 − 6y) = 0

9x2 + 5 (y2 − 6y + 9) = 45

x2 / 5 + (y − 3)2 / 9 = 1

∵ a2 < b2, so axis of the ellipse is on the y-axis.

At y-axis, put x = 0, so we can obtain the vertex.

Then 0 + 5y2 − 30y = 0

y = 0, y = 6

Therefore, tangents of vertex y = 0, y = 6.

Example 5: The equation of tangent and normal at point (3, 2) of ellipse 4x2 + 9y2 = 36 are _________.

Solution:

Given, equation of an ellipse is 4x2 + 9y2 = 36

Tangent at point (3, 2) is (3) * x / 9 + (−2) * y / 4 = 1 or x / 3 − y / 2 = 1

∴Normal is x / 2 + y / 3 = k and it passes through the point (3,2)

∴ 3 / 2 − 2 / 3 = k ⇒ k = 5 / 6

∴ Normal is, x / 2 + y / 3 = 5 / 6.

Example 6: Minimum area of the triangle by any tangent to the ellipse x2 / a2 + y2 / b2 = 1 with the coordinate axes is _____________.

Solution:

Equation of tangent at (a cos θ, b sin θ) is [x / a] cos θ + [y / b] sin θ = 1

P = (a / cos θ, 0)

Q = (0, b / sin θ)

Area of OPQ = 1 / 2 ∣(a / cos θ) (b / sin θ)∣ = ab / |sin 2θ|

(Area)min = ab

Example 7: The eccentric angles of the extremities of latus recta of the ellipse x2a2+y2b2=1\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 are given by

A)tan1(±aeb)B)tan1(±bea)C)tan1(±bae)D)tan1(±abe)A){{\tan }^{-1}}\left( \pm \frac{ae}{b} \right)\\ B){{\tan }^{-1}}\left( \pm \frac{be}{a} \right)\\ C){{\tan }^{-1}}\left( \pm \frac{b}{ae} \right)\\ D){{\tan }^{-1}}\left( \pm \frac{a}{be} \right)

Solution: 

Coordinates of any point on the ellipse x2a2+y2b2=1\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1

whose eccentric angle is θ\theta are (acosθ,bsinθ).(a\cos \theta ,\,\,b\sin \theta ).

The coordinates of the end points of latus recta are (ae,±b2a).\left( ae,\,\pm \frac{{{b}^{2}}}{a} \right). 

Because 

acosθ=ae and bsinθ=±b2atanθ=±baeθ=tan1(±bae).a\cos \theta =ae \text \ and \ b\sin \theta =\pm \frac{{{b}^{2}}}{a} \\ \tan \theta =\pm \frac{b}{ae}\Rightarrow \theta ={{\tan }^{-1}}\left( \pm \frac{b}{ae} \right).

Example 8: The length of the axes of the conic 9x2+4y26x+4y+1=0,9{{x}^{2}}+4{{y}^{2}}-6x+4y+1=0, are 

A)12, 9B)3, 25C)1, 23D)3,2A)\frac{1}{2},\ 9\\ B)3,\ \frac{2}{5}\\ C)1,\ \frac{2}{3}\\ D)3, 2

Solution:

 Given that, the equation of conic

9x2+4y26x+4y+1=0(3x1)2+(2y+1)2=1(x13)219+(y+1)212=1.9{{x}^{2}}+4{{y}^{2}}-6x+4y+1=0\\ {{(3x-1)}^{2}}+{{(2y+1)}^{2}}=1\\ \frac{{{\left( x-\frac{1}{3} \right)}^{2}}}{\frac{1}{9}}+\frac{{{(y+1)}^{2}}}{\frac{1}{2}}=1.

Here a=13,b=12;2a=23,2b=1.a=\frac{1}{3}, b=\frac{1}{2}; 2a=\frac{2}{3}, 2b=1.

Length of axes are (1,23)\left( 1,\,\frac{2}{3} \right)

Example 9: The coordinates of the foci of the ellipse 3x2+4y212x8y+4=03{{x}^{2}}+4{{y}^{2}}-12x-8y+4=0 are 

A) (1, 2), (3, 4)                               

B) (1, 4), (3, 1)

C) (1, 1), (3, 1)                               

D) (2, 3), (5, 4)

Solution:

3x212x+4y28y=43(x2)2+4(y1)2=12(x2)24+(y1)23=1X24+Y23=1e=134=12.3{{x}^{2}}-12x+4{{y}^{2}}-8y=-4\\ 3{{(x-2)}^{2}}+4{{(y-1)}^{2}}=12\\ \frac{{{(x-2)}^{2}}}{4}+\frac{{{(y-1)}^{2}}}{3}=1 \\ \frac{{{X}^{2}}}{4}+\frac{{{Y}^{2}}}{3}=1\\ e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}.

Foci are

(X=±2×12,Y=0)i.e., (x2=±1,y1=0)=(3,1) and (1,1).\left( X=\pm 2\times \frac{1}{2},\,Y=0 \right) \text i.e., \ (x-2=\pm 1,\,\,y-1=0) =(3,\,1) \text \ and \ (1,\,1).