Question

The sum of squares of the perpendiculars drawn from the points (0, 1) and (0, -1) to any tangent to a curve is 2. Then, the equation of the curve is

• A. $2y^{\prime }=c(x+2)$
• B. $y=c^{\prime }(x+1)$
• C. $2y=c^{\prime }(x+2)$
• D. $y{y}^{\prime }=c(x+1)$

Answer 1

The correct option is B $y=c^{\prime }(x+1)$

Equation of tangent to the curve at (x, y) is $Y-y=\frac{dy}{dx}(X-x)......\left(i\right)$
The length of perpendicular from (0, 1) to the curve is
$D_1=\left|\frac{1+x{y}^{\prime }-y}{\sqrt{1+(y^{\prime }{)}^2}}\right|......\left(ii\right)where{y}^{\prime }=\frac{dy}{dx}$
Then, length of perpendicular from (0, -1) to the curve is
$D_2=\left|\frac{-1+x{y}^{\prime }-y}{\sqrt{1+(y^{\prime }{)}^2}}\right|......\left(iii\right)\Rightarrow D_1^2+D_2^2=2\Rightarrow 2+2x^2{y}^{\prime 2}-2y^2-4xy{y}^{\prime }=2(1+y^{\prime 2})\Rightarrow x^2{y}^{\prime 2}+y^2-2xy{y}^{\prime }=y^{\prime 2}\Rightarrow {\left(\frac{dy}{dx}\right)}^2(x^2-1)-2xy\left(\frac{dy}{dx}\right)+y^2=0\Rightarrow \frac{dy}{dx}=\frac{2xy\pm \sqrt{4y^2}}{2(x^2-1)}=\frac{xy\pm y}{x^2-1}\Rightarrow \frac{dy}{y}=\frac{(x\pm 1)dx}{x^2-1}$
Case 1 Taking positive sign
$\frac{dy}{y}=\frac{dx}{x-1}\Rightarrow y=c(x-1)$
Case 2 Taking negative sign
$\frac{dy}{y}=\frac{x}{x+1}\Rightarrow y=c^{\prime }(x+1)$