Question

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

• A. $(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$
• B. $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$
• C. $(\frac{1}{\sqrt{7}},-\frac{1}{\sqrt{7}})$
• D. None of these

Answer 1

Answer: (A)

$(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$

$\begin{array}{c}3x^2+4y^2=1\\ Equationof\tan genttoellipseat\left(x_1,y_1\right)\\ 3x{x}_1+4y{y}_1=1\\ \frac{3x}{\sqrt{7}}+\frac{4y}{\sqrt{7}}=1\\ x_1=\frac{1}{\sqrt{7}}{y}_1=\frac{1}{\sqrt{7}}\\ \left(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}}\right)\\ Hence,\\ optionAiscorrectanswer.\end{array}$