Question

A tangent to $3x^2+4y^2=12$ is equally inclined with the coordinate axis. Then the perpendicular distance from the centre of the ellipse to this tangent is

• A. $\sqrt{\frac{7}{2}}$
• B. $\sqrt{\frac{5}{2}}$
• C. $\sqrt{\frac{9}{2}}$
• D. $\sqrt{\frac{11}{2}}$

Answer 1

The correct option is A $\sqrt{\frac{7}{2}}$

Let the equation of the tangent be,

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

Given equation of ellipse can be written as,

$\frac{x^2}{4}+\frac{y^2}{3}=1$

$\therefore$ $\frac{x}{2}\cos \theta +\frac{y}{\sqrt{3}}\sin \theta =1$

X-Intercept=$\frac{2}{\cos \theta }$

Y-Intercept=$\frac{\sqrt{3}}{\sin \theta }$

Given equal intercepts

So, on equating both the intercepts we get,

$\tan \theta =\frac{\sqrt{3}}{2}$

On putting the values of $\sin \theta$ and $\cos \theta$ in the tangent equation we get (values of $\sin \theta$ and $\cos \theta$ can be calculated from the triangle shown in fig.b )

$\therefore$ $\frac{x}{\sqrt{7}}+\frac{y}{\sqrt{7}}=1$

Now, center of the ellipse is (0,0)

Distance from the center=$\frac{|0+0-\sqrt{7}|}{\sqrt{1^2+1^2}}$

$\therefore$ Distance = $\sqrt{\frac{7}{2}}$