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.
• B.
• C.
• D.

Answer 1

The correct option is A

Let $P(x_1,y_1)$ be the point of contact.
The equation of the tangent at P is $x\frac{x_1}{16}+y\frac{y_1}{9}=1$
If this tangent passes through $(4,\sqrt{3})$, then $\frac{x_1}{4}+\frac{y_1}{3\sqrt{3}}=1\Rightarrow \frac{y_1}{3\sqrt{3}}=1-\frac{x_1}{4}\Rightarrow y_1=\frac{12\sqrt{3}-3\sqrt{3}{x}_1}{4}$
P lies on the ellipse $\Rightarrow \frac{x_1^2}{16}+\frac{y_1^2}{9}=1\Rightarrow \frac{x_1^2}{16}+\frac{{(12\sqrt{3}-3\sqrt{3}{x}_1)}^2}{144}=1\Rightarrow x_1^2+{(4\sqrt{3}-\sqrt{3}{x}_1)}^2=16$
$\Rightarrow x_1^2+48+3x_1^2-24x_1=16\Rightarrow 4x_1^2-24x_1+32=0\Rightarrow x_1^2-6x_1+8=0\Rightarrow x_1=2or4$
$x_1=2\Rightarrow y_1=3\frac{\sqrt{3}}{2},x_1=4\Rightarrow y_1=0$
$\therefore$ The required points are $(2,\frac{3\sqrt{3}}{2}),(4,0)$