Question

The equation(s) of normal(s) to the curve $3x^2-y^2=8$ which is (are) parallel to the line x + 3y = 4 is (are)

• A. $x+3y+8=0$
• B. $x+3y-8=0$
• C. $3x+y+8=0$
• D. $3x+y+8=0$

Answer 1

Answer: (A), (B)

$x+3y-8=0$

Let the reqiured normal(s) be drawn at a point P(a, b). The equation of the given curve is $3x^2-y^2=8$ ......[1]
Differentiating both sides w.r.t. x, we get
$6x-2y\frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3x}{y}\Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(a,b)}=\frac{3a}{b}\mathrm{Since},\mathrm{the}\mathrm{normal}\mathrm{at}\left(a,b\right)\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}x+3y=4.\mathrm{Therefore},\mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\left(a,b\right)=\mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{line}x+3y=4\Rightarrow -\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(a,b)}}=\frac{-1}{3}\Rightarrow \frac{-b}{3a}=-\frac{1}{3}\Rightarrow a=b..\left(2\right]\mathrm{Since},\left(a,b\right)\mathrm{lies}\mathrm{on}\left(1\right].\mathrm{Therefore},3a^2-b^2=8\Rightarrow 3a^2-a^2=8\left(\mathrm{From}\left(2\right]\right)\Rightarrow 2a^2=8\Rightarrow a^2=4\Rightarrow a=\pm 2\mathrm{If}a=2,\mathrm{then}b=2\mathrm{and}\mathrm{if}a=-2,\mathrm{then}b=-2\mathrm{Thus},\mathrm{the}\mathrm{co}-\mathrm{ordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{are}\left(2,2\right)\mathrm{and}\left(-2,-2\right).\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\left(2,2\right)\mathrm{is}y-2=-\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(a,b)}}\left(x-2\right)\Rightarrow y-2=-\frac{1}{3}\left(x-2\right)\Rightarrow x+3y-8=0\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\left(-2,-2\right)\mathrm{is}y+2=-\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(a,b)}}\left(x+2\right)\Rightarrow y+2=-\frac{1}{3}\left(x+2\right)\Rightarrow x+3y+8=0.$