Question

In the line $3x+4y=\sqrt{7}$ touches the ellipse $3x^2+4y^2=1$ , then the point of contact is-

• A. $\left(\frac{1}{\sqrt{7}},-\frac{1}{\sqrt{7}}\right)$
• B. $\left(\frac{2}{\sqrt{7}},\frac{1}{4\sqrt{7}}\right)$
• C. $\left(\frac{3}{\sqrt{7}},-\frac{1}{2\sqrt{7}}\right)$
• D. $\left(-\frac{1}{3\sqrt{7}},\frac{2}{\sqrt{7}}\right)$

Answer 1

The correct option is A $\left(\frac{1}{\sqrt{7}},-\frac{1}{\sqrt{7}}\right)$

In the line $3x+4y=\sqrt{7}$ touches the ellipse

$3x^2+4y^2=1$, then the point of contact

We know that

the condition for a straight line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2{m}^2+b^2$

from the point of contact

$=(\frac{-a^{2}m}{c},\frac{b^2}{c})$ & \left (\dfrac {a^2 m}{c}, \dfrac {-b^2}{c}\right)$

$\therefore 3x^2+4y^2=1$

$\Rightarrow \frac{x^2}{{\left(\frac{1}{\sqrt{3}}\right)}^2}+\frac{y^2}{{\left(\frac{1}{2}\right)}^2}=1$

$\therefore a=\frac{1}{\sqrt{3}}b=\frac{1}{2}$

$\therefore$ point of contact is

$=(\frac{-{(+\frac{1}{\sqrt{3}})}^2\left(\frac{-3}{2}\right)}{\frac{\sqrt{7}}{4}},\frac{{\left(\frac{1}{2}\right)}^2}{\frac{\sqrt{7}}{4}})$ & $(\frac{{\left(\frac{1}{\sqrt{3}}\right)}^2\left(\frac{-3}{2}\right)}{\frac{\sqrt{7}}{4}},\frac{{-\left(\frac{1}{2}\right)}^2}{\frac{\sqrt{7}}{4}})$

$=[-\frac{\left(\frac{1}{3}\right)\left(\frac{-3}{4}\right)}{\frac{\sqrt{7}}{4}},\frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}}]$ & $[\frac{\frac{1}{3}\times \frac{3}{4}}{\frac{\sqrt{7}}{4}},\frac{-\frac{1}{4}}{\frac{\sqrt{7}}{4}}]$

$=[\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}}]$ & $[\frac{1}{\sqrt{7}},-\frac{1}{\sqrt{7}}]$

$3x+4y=\sqrt{7}$

$4y=-3x+\sqrt{7}$

$y=\left(\frac{-3}{4}\right)x+\frac{\sqrt{7}}{4}$

$\therefore m=\frac{-3}{4}c=\frac{\sqrt{7}}{4}$