Question

Let the two foci of an ellipse be $(-1,0)$ and $(3,4)$ and the foot of perpendicular from the focus $(3,4)$ upon a tangent to the ellipse be $(4,6)$
The length of the semi-minor axis of the ellipse is

• A. $1$
• B. $2\sqrt{2}$
• C. $\sqrt{17}$
• D. $4\sqrt{3}$

Answer 1

The correct option is D $4\sqrt{3}$

Refer to the figure.

Slope of line SP is given by,

$m_1=\frac{y_2-y_1}{x_2-x_1}$

$\therefore m_1=\frac{6-4}{4-3}$

$\therefore m_1=\frac{2}{1}=2$

Now, tangent through point P is perpendicular to line SP.

Thus, slope of tangent is $\frac{-1}{2}$

Thus, equation of tangent will be given by,

$y-y_1=m(x-x_1)$

$\therefore y-6=\frac{-1}{2}(x-4)$

$\therefore 2(y-6)=-(x-4)$

$\therefore 2y-12=-x+4$

$\therefore x+2y-16=0$

$\therefore x+2y=16$

Dividing both sides by 16, we get,

$\frac{x}{16}+\frac{2y}{16}=\frac{16}{16}$

$\therefore \frac{x}{16}+\frac{y}{8}=1$ (1)

We know that equation of tangent to the ellipse through any point on the ellipse is given by,

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

$\therefore \frac{4x}{a^2}+\frac{6y}{b^2}=1$ (2)

Comparing coefficients of x and y in equations (1) and (2),

$\therefore \frac{4}{a^2}=\frac{1}{16}$

$\therefore a^2=64$

$\therefore a=8$

Similarly, $\frac{6}{b^2}=\frac{1}{8}$

$\therefore b^2=48$

$\therefore b=\sqrt{48}$

$\therefore b=4\sqrt{3}$

Thus, length of the semi-minor axis of ellipse is $4\sqrt{3}$