Two ellipses have a common focus ; two radii vectores, one to each ellipse, are drawn from the focus at right angles to one another and tangents are drawn at their extremities ; prove that these tangents meet on a fixed conic, and find when it is a parabola.

Open in App

Solution

The equation to the ellipses having the same focus may be taken as
$\frac{l}{r} = 1 - e cos θ . . . . .1$
and $\frac{l_{1}}{r} = 1 - e_{1} cos (θ - α) . . . . .2$
Let ' $β$ ' be the vectorial angle of any point P on $1$ ; then the vectorial angle of the point Q on $2$ will be $(90 + β)$
Equation of tangent to $1$ at point P will be
$\frac{l}{r} = cos (θ - β) - e cos θ . . . .3$
Equation of tangent to $2$ at point Q will be
$\frac{l_{1}}{r} = cos (θ - 90 - β) - e_{1} cos (θ - α)$
or $\frac{l_{1}}{r} = sin (θ - β) - e_{1} cos (θ - α) . . . .4$
To get the locus of the required point, we have to eliminate $β$ from $3$ and $4$ .
As we have
$\frac{l}{r} + e cos θ = cos (θ - β)$ .......... by $3$
And $\frac{l_{1}}{r} + e_{1} cos (θ - α) = sin (θ - β)$ ......... by $4$
Squaring and adding the two equations, we get
${(\frac{l}{r} + e cos θ)}^{2} + {\frac{l_{1}}{r} + e_{1} cos (θ - α)}^{2} = 1$
Which is the locus of the point of intersection of the tangents and clearly it is a fixed conic
Changing the above equation to Cartesian we get
${(l + e x)}^{2} + {(l_{1} + e_{1} x cos α + e_{1} y sin α)}^{2} = x^{2} + y^{2}$
If it is a parabola, the terms of the second degree must form a perfect square, hence the required condition is
$l - e^{2} - e_{1}^{2} + e^{2} e_{1}^{2} {sin}^{2} α = 0$

Suggest Corrections

Similar questions

2 {sin}^{- 1} \frac{e}{\sqrt{2}} .

y^{2} = 4 a (x + a)

y^{2} = 4 a^{'} (x + a^{'})

x + a + a^{'} = 0

Join BYJU'S Learning Program