Question

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

• A. $(\frac{\sqrt{7}}{3},0)$
• B. $(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$
• C. $\left(0,\frac{\sqrt{7}}{4}\right)$
• D. undefined

Answer 1

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

Let the given line touches the ellipse
at $\left(x_1,y_1\right)$,
then the equation of line is
$3x_{1}x+4y_{1}y=1$
comparing it to given equation
$3x+4y=\sqrt{7}\frac{3}{\sqrt{7}}x+\frac{4}{\sqrt{7}}y=1$
We have $3x_1=\frac{3}{\sqrt{7}}\Rightarrow x_1=\frac{1}{\sqrt{7}}4y_1=\frac{4}{\sqrt{7}}\Rightarrow y_1=\frac{1}{\sqrt{7}}$
So, the point of contact is $\left(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}}\right)$