Question

The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the x-axis and passing through$\left( {2,1} \right)$, is:
A.${{\rm{x}}^2} + {{\rm{y}}^2} - {\rm{x}} = 0$
B.$4{{\rm{x}}^2} + 2{{\rm{y}}^2} - 9{\rm{y}} = 0$
C.$2{{\rm{x}}^2} + 4{{\rm{y}}^2} - 9{\rm{x}} = 0$
D.$4{{\rm{x}}^2} + 2{{\rm{y}}^2} - 9{\rm{x}} = 0$

Answer 1

Here we have to use the equation of the normal to find the equation of the x-intercept. Then an equation will form with the statement given in the question. The equation we will get is in the deferential form then we have to integrate that equation to get the actual equation of the curve. Then the equation we will get is with a constant and we know that the curve is passing through the point given and the equation of the curve will satisfy at this point. Then by this, we will get the value of the constant and then we will get the actual equation of the curve.

Complete step by step solution:
Firstly we will write the equation of the normal passing through the point $({\rm{x,y}})$. Therefore the equation of the normal at $({\rm{x,y}})$ is
${\rm{Y}} - {\rm{y}} = - \dfrac{{{\rm{dx}}}}{{{\rm{dy}}}}({\rm{X}} - {\rm{x)}}$
It is given that the intercept of the normal on the x-axis. Therefore putting the value of Y equals to 0, we get
X-intercept ${\rm{ = y}}\dfrac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{x}}$
Now according to the question, the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the x-axis and passing through $\left( {2,1} \right)$. So from this, we will get an equation
$\Rightarrow {{\rm{y}}^2} = 2{\rm{x}}\left( {{\rm{y}}\dfrac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{x}}} \right)$
Now we have to simplify this equation, we get
$\Rightarrow {{\rm{y}}^2} = 2{\rm{xy}}\dfrac{{{\rm{dy}}}}{{{\rm{dx}}}} + 2{{\rm{x}}^2}$
$\Rightarrow \dfrac{{{\rm{dy}}}}{{{\rm{dx}}}} = \dfrac{{{{\rm{y}}^2} - 2{{\rm{x}}^2}}}{{2{\rm{xy}}}}$………….. (1)
Now to solve this equation we have to put ${\rm{y}} = {\rm{vx}}$ and by its differentiation we get$\dfrac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{v + x}}\dfrac{{{\rm{dv}}}}{{{\rm{dx}}}}$
Therefore we get
$\Rightarrow {\rm{v + x}}\dfrac{{{\rm{dv}}}}{{{\rm{dx}}}} = \dfrac{{{{\rm{v}}^2} - 2}}{{2{\rm{v}}}}$
$\Rightarrow {\rm{x}}\dfrac{{{\rm{dv}}}}{{{\rm{dx}}}} = \dfrac{{{{\rm{v}}^2} - 2}}{{2{\rm{v}}}} - {\rm{v}} = \dfrac{{ - (2 + {{\rm{v}}^2})}}{{2{\rm{v}}}}$
Now by taking the same variable terms on the same side, we get
$\Rightarrow \dfrac{{2{\rm{v}}}}{{2 + {{\rm{v}}^2}}}{\rm{dv}} = - \dfrac{{{\rm{dx}}}}{{\rm{x}}}$
Now we have to integrate both sides to get the final equation, we get
$\Rightarrow \int {\dfrac{{2{\rm{v}}}}{{2 + {{\rm{v}}^2}}}} {\rm{dv}} = - \int {\dfrac{{{\rm{dx}}}}{{\rm{x}}}}$
$\Rightarrow \log (2 + {{\rm{v}}^2}) = - \log {\rm{x + logk}}$
Now by simplifying the equation we get
$\Rightarrow \log (2 + {{\rm{v}}^2}) + \log {\rm{x = logk}}$
$\Rightarrow \log {\rm{x}}(2 + {{\rm{v}}^2}){\rm{ = logk}}$
$\Rightarrow {\rm{x}}(2 + {{\rm{v}}^2}) = {\rm{k}}$
Now we have to replace the variable v with ( y/x) as we know that ${\rm{y}} = {\rm{vx}}$. Therefore, we get
$\Rightarrow {\rm{x}}(2 + \dfrac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}) = {\rm{k}}$
$\Rightarrow {{\rm{y}}^2} + 2{{\rm{x}}^2} - {\rm{kx = 0}}$
And it is given in the question that the curve is passing through the point$\left( {2,1} \right)$. Therefore by putting this value in the equation we get the value of k.
$\Rightarrow {1^2} + 2\left( {{2^2}} \right) - {\rm{k2 = 0}}$
$\Rightarrow {\rm{k = }}\dfrac{9}{2}$
Therefore the equation becomes $4{{\rm{x}}^2} + 2{{\rm{y}}^2} - 9{\rm{x}} = 0$
Hence, option D is the correct option.

Note: We should know that the equation of the normal value of Y is zero when its intercept is the x-axis and the value of X is zero when the intercept of normal is the y-axis. Intercept is the point of intersection of a curve or a line on the x-axis or y-axis. The abscissa is the distance of a point on the x-axis i.e. which is measured parallel to the x-axis or we can say that it is the x coordinate of the point. If any curve passing through a point then the value of the coordinate of the point always satisfies the equation of the curve. We should note that while integrating the equation should not be complex.