JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

Equation of Tangent to Ellipse

An ellipse is a curve in the plane which surrounds the two focal points such that the distances to the focal point remain constant for each point on the curve. A circle is said to be a special type of 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] represents 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.

  • The 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 the 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

The 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 the axis of the ellipse is on the y-axis.

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

The 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:

The 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

\(\begin{array}{l}\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\end{array} \)
are given by

\(\begin{array}{l}(A)\ {{\tan }^{-1}}\left( \pm \frac{ae}{b} \right)\\ (B)\ {{\tan }^{-1}}\left( \pm \frac{be}{a} \right)\\ (C)\ {{\tan }^{-1}}\left( \pm \frac{b}{ae} \right)\\ (D)\ {{\tan }^{-1}}\left( \pm \frac{a}{be} \right)\end{array} \)

Solution: 

Coordinates of any point on the ellipse

\(\begin{array}{l}\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\end{array} \)

whose eccentric angle is θ are (a cos θ, b sin θ). 

 

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

Because 

\(\begin{array}{l}a\cos \theta =ae\ \text{and}\ \ b\sin \theta =\pm \frac{{{b}^{2}}}{a} \\ \tan \theta =\pm \frac{b}{ae}\Rightarrow \theta ={{\tan }^{-1}}\left( \pm \frac{b}{ae} \right).\end{array} \)

Thus, the answer is option C.

Example 8: The length of the axes of the conic

\(\begin{array}{l}9{{x}^{2}}+4{{y}^{2}}-6x+4y+1=0,\end{array} \)
are 

\(\begin{array}{l}(A)\ \frac{1}{2},\ 9\\ (B)\ 3,\ \frac{2}{5}\\ (C)\ 1,\ \frac{2}{3}\\ (D)\ 3, 2\end{array} \)

Solution:

 Given that, the equation of the conic

\(\begin{array}{l}9{{x}^{2}}+4{{y}^{2}}-6x+4y+1=0\\ {{(3x-1)}^{2}}+{{(2y+1)}^{2}}=1\\ \frac{{{\left( x-\frac{1}{3} \right)}^{2}}}{\frac{1}{9}}+\frac{{{(y+1)}^{2}}}{\frac{1}{2}}=1.\end{array} \)

Here

\(\begin{array}{l}a=\frac{1}{3}, b=\frac{1}{2}; 2a=\frac{2}{3}, 2b=1.\end{array} \)

The length of the axes of the conic are 1, 2/3.

Thus, the answer is option C.

Example 9: The coordinates of the foci of the ellipse

\(\begin{array}{l}3{{x}^{2}}+4{{y}^{2}}-12x-8y+4=0\end{array} \)
are 

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

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

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

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

Solution:

\(\begin{array}{l}3{{x}^{2}}-12x+4{{y}^{2}}-8y=-4\\ 3{{(x-2)}^{2}}+4{{(y-1)}^{2}}=12\\ \frac{{{(x-2)}^{2}}}{4}+\frac{{{(y-1)}^{2}}}{3}=1 \\ \frac{{{X}^{2}}}{4}+\frac{{{Y}^{2}}}{3}=1\\ e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}.\end{array} \)

Foci are

\(\begin{array}{l}\left( X=\pm 2\times \frac{1}{2},\,Y=0 \right) \text i.e., \ (x-2=\pm 1,\,\,y-1=0) =(3,\,1)\ \text{and}\ \ (1,\,1).\end{array} \)

Thus, the answer is option C.

Definitions of an Ellipse

Visualising Ellipses

Frequently Asked Questions

Q1

What do you mean by a tangent to an ellipse?

A tangent to an ellipse is a line that intersects the ellipse at one point.

Q2

Give the equation for the 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 given by (xx1/a2) + (yy1/b2) = 1.

Q3

What is the equation of normal to an ellipse?

The equation of the normal to the ellipse (x2/a2) + (y2/b2) = 1 at the point (x1, y1) is given by (a2x/x1) – (b2y/y1) = a2 – b2.

Q4

How many tangents can be drawn to an ellipse from an external point?

Two tangents can be drawn to an ellipse from an external point.

Test your Knowledge on equation of tangent to ellipse

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*