Question

Tangents are drawn through the point $(4,\sqrt{3})$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$, the points at which these tangents touch the ellipse are:

• A. $(2,\frac{3\sqrt{3}}{2})$
• B. $(2,-\frac{3\sqrt{3}}{2})$
• C. $(-4,0)$
• D. $(0,-4)$

Answer 1

The correct option is A $(2,\frac{3\sqrt{3}}{2})$

Given ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$

Let $(x_0,y_0)$ be a point on ellipse at which tangent meets.

Equation of tangent at $(x_0,y_0)$ is:

$y-y_0=\frac{-9x_0}{16y_0}(x-x_0)$

Given point $(4,\sqrt{3})$ lie on the line

$\to \sqrt{3}-y_0=\frac{-9x_0}{16y_0}(4-x_0)$

And $(x_0,y_0)$ lie on ellipse

$\frac{x_0^2}{16}+\frac{y_0^2}{9}=1$

$\to y_0=3\sqrt{1-\frac{x_0^2}{16}}$

Simplifying line equation :

$16\sqrt{3}{y}_0-16y_0^2=-36x_0+9x_0^2$

Substituting $y_0$ in above equation ;

On simplifying we get,

$\sqrt{3}\sqrt{16-x_0^2}=12-3x_0$

Squaring on both sides ;

$\Rightarrow 3(16-x_0^2)=9(16+x_0^2-8x_0)$

$\Rightarrow x_0^2-6x_0+8=0$

$\Rightarrow (x_0-4)(x_0-2)=0$

$\Rightarrow x_0=4$ or $x_0=2$

For $x_0=4$ $\to y_0=0$

For$x_0=2$ $\to y_0=\frac{3\sqrt{3}}{2}$