Name: Tangent & Normal to Ellipse
Uploaded: 2019-07-19
Duration: 4 min 30 s
Description: We have studied the tangent to a circle. A circle is a special case of ellipses. Learn more about a tangent to an ellipse by watching this video.

Question

A line of fixed length

(a + b)

moves so that its ends are always on two fixed perpendicular straight lines. Prove that a marked point on the line, which divides this line into portions of length

' a'

and

' b'

describes an ellipse and also find the eccentricity of the ellipse when

a = 8

,

b = 12

.

Accepted Answer

Let the two perpendicular lines $x = 0$ and $y = 0$

Let the coordinates of point be $(x, 0), (0, y)$

By section formula $h = \frac{a x + b (0)}{a + b}, k = \frac{a (0) + b y}{a + b}$

By Pythagoras theorem

$(a + b)^{2} = x^{2} + y^{2} \dots (1)$

From above, $x = \frac{(a + b) h}{a}, y = \frac{(a + b) k}{b}$

$∴ x^{2} = \frac{(a + b)^{2} h^{2}}{a^{2}}$ , $y^{2} = \frac{(a^{2} + b^{2}) k^{2}}{b^{2}}$

Substitute these values in equation $(1)$

$∴ \frac{(a + b)^{2} h^{2}}{a^{2}} + \frac{(a + b)^{2} k^{2}}{b^{2}} = (a + b)^{2}$

$∴ \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} = 1$ is standard ellipse.

Hence, proved that a marked point $(h, k)$ on the line which divides this line into portions of length $' a'$ and $b$ .

Now we have to find eccentricity of ellipse $e = \sqrt{1 - \frac{a^{2}}{b^{2}}}$

Hence, $e = \sqrt{1 - \frac{64}{144}} = \sqrt{\frac{5}{9}}$

$e = \frac{\sqrt{5}}{3}$

Important Questions on Ellipse

Learn and Prepare for any exam you want !

A line of fixed length a+b moves so that its ends are always on two fixed perpendicular straight lines. Prove that a marked point on the line, which divides this line into portions of length 'a' and 'b' describes an ellipse and also find the eccentricity of the ellipse when a=8, b=12.

Important Questions on Ellipse

Learn and Prepare for any exam you want !