Question

If the line $3x+4y=7$ is a normal at a point $P=(x_1,y_1)$ of the hyperbola $3x^2-4y^2=1,$ then the distance of $P$ from the origin is

• A. $\frac{\sqrt{319}}{12}$
• B. $\frac{\sqrt{337}}{12}$
• C. $\frac{\sqrt{423}}{12}$
• D. $\frac{\sqrt{527}}{12}$

Answer 1

The correct option is B $\frac{\sqrt{337}}{12}$

For the hyperbola $3x^2-4y^2=1$,
Slope of the tangent ,
$6x-8y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{3x}{4y}$
So, slope of normal at point$(x_1,y_1)$,
$m_n=-\frac{4y_1}{3x_1}=-\frac{3}{4}$
$\Rightarrow 9x_1-16y_1=0\cdots \left(1\right)$
Also point $(x_1,y_1)$ lies on the line $3x+4y=7$
$\Rightarrow 3x_1+4y_1=7\cdots \left(2\right)$
solving $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow x_1=\frac{4}{3},y_1=\frac{3}{4}$
Distance from the origin
$\Rightarrow OP=\sqrt{{\left(\frac{4}{3}\right)}^2+{\left(\frac{3}{4}\right)}^2}=\frac{\sqrt{337}}{12}$