Question

The line $3x+4y=\sqrt{7}$ touches the ellipse $3x^2+4y^2=1$ at the point

• A. $(\sqrt{7},\sqrt{7})$
• B. $(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$
• C. $(\frac{1}{\sqrt{7}},-\frac{1}{\sqrt{7}})$
• D. $(-\sqrt{7},\sqrt{7})$

Answer 1

The correct option is B $(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$

Let the given line touches the ellipse at point $P\left(\theta \right)=(\frac{\cos \theta }{\sqrt{3}},\frac{\sin \theta }{2})$

The equation of the tangent at $P$ is
$\sqrt{3}x\cos \theta +2y\sin \theta =1\dots \left(i\right)$

Comparing equation $\left(i\right)$ with the given equation of the line $3x+4y=\sqrt{7},$
we get
$\Rightarrow \frac{\cos \theta }{\sqrt{3}}=\frac{\sin \theta }{2}=\frac{1}{\sqrt{7}}$

$\therefore$ Point of contact is $(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$

Alternate Solution:
For the given ellipse $a^2=1/3,b^2=1/4$
Given tangent is
$3x+4y=\sqrt{7}\Rightarrow y=-\frac{3}{4}x+\frac{\sqrt{7}}{4}$
$\therefore m=-\frac{3}{4},c=\frac{\sqrt{7}}{4}$
Using slope form the point of contact will be
$(-\frac{a^{2}m}{c},\frac{b^2}{c})\equiv (\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$