An ellipse is drawn with major and minor axes | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$2a = 10 \Rightarrow a = 5 ; 2b = 8 \Rightarrow b = 4$ 
$e^2 = 1 - \frac{{16}}{{25}}\,$=$\frac{9}{{25}}\, \Rightarrow e = \frac{3}{5}\,$&nb

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$2a = 10 \Rightarrow a = 5 ; 2b = 8 \Rightarrow b = 4$ 
$e^2 = 1 - \frac{{16}}{{25}}\,$=$\frac{9}{{25}}\, \Rightarrow e = \frac{3}{5}\,$ 
Focus = $(3, 0)$ 
Let the circle touches the ellipse at $ P $ and $Q$. Consider a tangent (to both circle and ellipse) at $P$. Let $F$(one focus) be the centre of the circle and other focus be $G.$  A ray from $F$  to $ P$ must retrace its path (normal to the circle). But the reflection propety the ray $FP$  must be reflected along $PG$. This is possible only if $ P, F $ and $G $ are collinear. Thus $P$ must be the end of the major axis.Hence $ r = a - ae = 5 - 3 = 2$ 
  alternately normal to an ellipse at $P$  must pass through the centre $(3, 0)$  of the circle 
$\frac{{ax}}{{\cos 	heta }}\,\, - \,\,\frac{{by}}{{\sin 	heta }}\,\, = \,\,{a^2} - {b^2} $ $\Rightarrow \frac{{5x}}{{\cos 	heta }}\,\, - \,\,\frac{{4y}}{{\sin 	heta }}\,\, = \,\,9$ $\left( {	heta \, 
e \,0\,\,or\,\frac{\pi }{2}\,} ight)\,$
$\frac{{15}}{{\cos 	heta }}\,\, - \,\,0\, = 9\, \Rightarrow \cos 	heta \, = \,\frac{{15}}{9}\,$  $\Rightarrow$ which is not possible $\Rightarrow 	heta\ = 0 or \pi /2$ 
but $	heta \pi /2 \Rightarrow 	heta = 0$
Hence $P (5, 0)$ i.e. end of major axis

An ellipse is drawn with major and minor axes of lengths $10 $ and $8$ respectively. Using one focus as centre, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. The radius of the circle is

Similar Questions

If tangents are drawn from the point ($2 + 13cos\theta , 3 + 13sin\theta $) to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1,$ then angle between them, is

If the tangents on the ellipse $4x^2 + y^2 = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^2$ is equal to

Find the equation for the ellipse that satisfies the given conditions: Vertices $(0,\,\pm 13),$ foci $(0,\,±5)$.

The eccentricity of ellipse $(x-3)^2 + (y -4)^2 = \frac{y^2}{9} +16 ,$ is -

The locus of point of intersection of two perpendicular tangent of the ellipse $\frac{{{x^2}}}{{{9}}} + \frac{{{y^2}}}{{{4}}} = 1$ is :-