Question

The lengths and the equations of the focal radii drawn to the point $(4\sqrt{3},5)$ on the ellipse $25x^2+16y^{{}^2}=1600$, are

• A. $13,4\sqrt{3y}-11x+24\sqrt{3}=0$
• B. $7,x+4\sqrt{3y}-24\sqrt{3}=0$
• C. Both (A) and (B)
• D. None of these

Answer 1

The correct option is C Both (A) and (B)

The equation of the ellipse is

$25x^2+16y^2=1600$

or $\frac{x^2}{64}+\frac{y^2}{100}=1$

Here $b>a$

$a^2=64,b^2=100$

$a^2=b^2(1-e^2)$

$\therefore 64=100(1-e^2)$

$⟹e=\frac{3}{5}$

Let $P(x_1,y_1)=(4\sqrt{3},5)$

be a point on the ellipse then SP and S'P are the focal radii

$\therefore SP=b-e{y}_1$ and

$S^{\prime }P=b+e{y}_1$

$\therefore SP=10-\frac{3}{5}\times 5$ and

$S^{\prime }P=10+\frac{3}{5}\times 5$

$⟹SP=7,S^{\prime }P=13$

Also $S$ be $(0,be)$

$i.e(0,10\times \frac{3}{5})$.

we get $(0,6)$ and

$S^{\prime }(0,-be)i.e(0,-10\times \frac{3}{5})$

$\therefore S^{\prime }(0,-6)$

Equation of SP is $y-5=\frac{(6-5)}{(0-4\sqrt{3})}\times (x-4\sqrt{3})-4\sqrt{(}3y)+20\sqrt{3}$

$⟹x+4\sqrt{3}y-24\sqrt{3}=0$

and Equation of S'P is $y-5=\frac{(-6-5)}{(0-4\sqrt{3})}\times (x-4\sqrt{3})$

$⟹-4\sqrt{3}y+20\sqrt{3}=-11x+44\sqrt{3}$

Thus, $11x-4\sqrt{3}y-24\sqrt{3}=0$