Question

$\mathrm{The}\mathrm{equation}\left(s\right)\mathrm{of}\mathrm{the}\mathrm{normal}\left(s\right)\mathrm{to}\mathrm{the}\mathrm{hyperbola}$ $3x^2-y^2=1\mathrm{having}\mathrm{slope}\frac{1}{3}$ $\mathrm{is}/\mathrm{are}:$

• A. $\sqrt{2}(x-3y)=4$
• B. $\sqrt{2}(x-3y)=-4$
• C. $\sqrt{2}(x-3y)=8$
• D. $\sqrt{2}(x-3y)=-8$

Answer 1

Answer: (A), (B)

$\sqrt{2}(x-3y)=-4$

$\mathrm{Differentiating}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{hyperbola}w.r.tx,\mathrm{we}\mathrm{get}:$
$6x-2y\frac{\mathrm{dy}}{\mathrm{dx}}=0\mathrm{or}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3x}{y}\mathrm{Let}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{curve}\mathrm{where}\mathrm{the}\mathrm{normal}\mathrm{has}\mathrm{slope}\frac{1}{3}\mathrm{be}P(x_1,y_1).\mathrm{Hence},-\frac{\mathrm{dx}}{\mathrm{dy}}=-\frac{y_1}{3x_1}=\frac{1}{3}\mathrm{or}{y}_1=-x_1\mathrm{Also},P\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{hyperbola}\therefore 3{x_1}^2-{y_1}^2=1\Rightarrow 3{x_1}^2-{x_1}^2=1\Rightarrow x_1=\pm \frac{1}{\sqrt{2}}\mathrm{Also},y_1=\mp \frac{1}{\sqrt{2}}\mathrm{Hence},\mathrm{the}\mathrm{points}\mathrm{on}\mathrm{the}\mathrm{curve}\mathrm{are}(\pm \frac{1}{\sqrt{2}},\mp \frac{1}{\sqrt{2})}\mathrm{Hence},\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{normals}\mathrm{are}:y-\frac{1}{\sqrt{2}}=\frac{1}{3}(x+\frac{1}{\sqrt{2}})\mathrm{and}y+\frac{1}{\sqrt{2}}=\frac{1}{3}(x-\frac{1}{\sqrt{2}})\Rightarrow \sqrt{2}(x-3y)=4\mathrm{and}\sqrt{2}(x-3y)=-4$