Question

Show that the ellipse (x^2)/(a^2)+2y^2 = 1 ?

Show that the ellipse `(x^2)/(a^2)+2y^2`= 1 and the hyperbola `(x^2)/(a^2-1)` `-2y^2`= 1 intersect at right angles.

Answer 1

Please see below.

Explanation:

Let the point of intersection at which we are seeking tangent be `(x_0,y_0)`. As it satisfies both the equations `x^2/a^2+2y^2=1` and `x^2/(a^2-1)-2y^2=1`, we have

`x_0^2/a^2+2y_0^2=1`

`x_0^2/(a^2-1)-2y_0^2=1`

Now subtracting latter from former, we get

`x_0^2(1/a^2-1/(a^2-1))+4y_0^2=0`

or `x_0^2/(a^2(a^2-1))=4y_0^2` or `x_0^2=4y_0^2a^2(a^2-1)` ........(A)

Using polarization tangents at ellipse `x^2/a^2+2y^2=1` and hyperbola `x^2/(a^2-1)-2y^2=1` are

`(x x_0)/a^2+2yy_0=1` and `(x x_0)/(a^2-1)-2yy_0=1` respectively

and their slopes are `-x_0/(2y_0a^2)` and `x_0/(2y_0(a^2-1))`

and product of slopes is `-x_0^2/(4y_0^2a^2(a^2-1))`

and using (A), product of slopes is `-1`

and hence the ellipse and the hyperbola intersect at right angles.