Question

The number of tangent(s) which can be drawn to the ellipse $16x^2+25y^2=400$, such that sum of perpendicular distance from the foci to the tangent is 8 , is

Answer 1

0

Answer 2

The equation of the ellipse is given by $16x^2 + 25y^2 = 400$.
Dividing by 400, we get the standard form:
$\frac{16x^2}{400} + \frac{25y^2}{400} = \frac{400}{400}$
$\frac{x^2}{25} + \frac{y^2}{16} = 1$

This is an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 25$ and $b^2 = 16$.
So, $a = 5$ and $b = 4$.
Since $a > b$, the major axis is along the x-axis.
The distance from the center to the foci is $c$, where $c^2 = a^2 - b^2$.
$c^2 = 25 - 16 = 9$
$c = 3$.
The foci of the ellipse are $F_1 = (-c, 0) = (-3, 0)$ and $F_2 = (c, 0) = (3, 0)$.

Let the equation of a tangent to the ellipse be $L$. Let $p_1$ be the perpendicular distance from the focus $F_1$ to the tangent $L$, and $p_2$ be the perpendicular distance from the focus $F_2$ to the tangent $L$.
The problem states that the sum of the perpendicular distances from the foci to the tangent is 8, i.e., $p_1 + p_2 = 8$.

A well-known property of the ellipse is that the sum of the perpendicular distances from the foci to any tangent is equal to $2a$, where $a$ is the semi-major axis length.
In this case, the semi-major axis length is $a = 5$.
According to the property, the sum of the perpendicular distances from the foci to any tangent is $2a = 2 \times 5 = 10$.

The problem requires the sum of the perpendicular distances from the foci to the tangent to be 8.
We have calculated that the sum of the perpendicular distances from the foci to any tangent of this ellipse is 10.
The required sum (8) is not equal to the actual sum for any tangent (10).
Therefore, there is no tangent to the given ellipse such that the sum of the perpendicular distances from the foci to the tangent is 8.

The number of such tangents is 0.