Question

The equation of the curve in which the portion of the tangent included between the coordinate axes is bisected at the point of contact is:
A. A parabola
B. An ellipse
C. A hyperbola
D. A circle

Answer 1

Hint: We can take the point of contact as$P\left( {x,y} \right)$. The slope of the tangent to the curve is given by, $\dfrac{{dy}}{{dx}}$ . Using the slope and point, we can get the equation of the tangent using the equation$\left( {y - {y_0}} \right) = \dfrac{{dy}}{{dx}}\left( {x - {x_0}} \right)$. Then we can find the points at which the tangent meets the x axis and y axis by substituting $y = 0$and $x = 0$in the equation of the tangent. Then we can apply midpoint formula $\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$to obtain a differential equation. We can solve them by integrating, to get the equation of the required curve.

Complete step by step Answer:

Let us take $P\left( {x,y} \right)$as the point of contact of the tangent and the curve. We know that the slope of the tangent to the curve is given by$\dfrac{{dy}}{{dx}}$.
The equation of a line passing through point $\left( {{x_0},{y_0}} \right)$and slope m is given by, $\left( {y - {y_0}} \right) = m\left( {x - {x_0}} \right)$
To get the equation of the tangent, we can substitute the point and the slope of the tangent.
$ \Rightarrow \left( {Y - y} \right) = \dfrac{{dy}}{{dx}}\left( {X - x} \right)$
Now we can find coordinates of the points where the tangent meets the axes. As these points lie on the tangent, it satisfied the equation of the tangent.
The y-intercept is given by,
For a point on y-axis,$X = 0$. We can substitute this in the equation of the tangent.
$ \Rightarrow \left( {Y - y} \right) = \dfrac{{dy}}{{dx}}\left( {0 - x} \right)$
On simplification, we get,
$ \Rightarrow Y - y = - x\dfrac{{dy}}{{dx}}$
$ \Rightarrow Y = y - x\dfrac{{dy}}{{dx}}$
Therefore, the point is A $\left( {0,y - x\dfrac{{dy}}{{dx}}} \right)$
For the x-intercept, y coordinate will be zero.
$ \Rightarrow Y = 0$.
We can substitute this in the equation on the tangent to obtain the x coordinate.
\[ \Rightarrow \left( {0 - y} \right) = \dfrac{{dy}}{{dx}}\left( {X - x} \right)\]
On simplification, we get,
\[ \Rightarrow \left( {X - x} \right) = - y\dfrac{{dx}}{{dy}}\]
\[ \Rightarrow X = x - y\dfrac{{dx}}{{dy}}\]
Therefore, the point is B $\left( {x - y\dfrac{{dx}}{{dy}},0} \right)$
From the question, the point of contact is the bisector of the tangent’s part connecting the 2 axes.
By mid-point formula, the mid-point of the line segment joining the points A $\left( {{x_1},{y_1}} \right)$and B$\left( {{x_2},{y_2}} \right)$is
given by, $\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Substituting the values for x coordinates of A and B, we get,
$x = \dfrac{{0 + x - y\dfrac{{dx}}{{dy}}}}{2}$
On multiplying the equation by 2 we get,
$ \Rightarrow 2x = x - y\dfrac{{dx}}{{dy}}$
On simplification we get,
$ \Rightarrow x = - y\dfrac{{dx}}{{dy}}$
On separating variables we get,
\[ \Rightarrow \dfrac{{dy}}{y} = - \dfrac{{dx}}{x}\]
We can integrate on both sides
\[ \Rightarrow \int {\dfrac{{dy}}{y}} = - \int {\dfrac{{dx}}{x}} \]
We know that $\int {\dfrac{1}{x}dx} = \log x$.
\[ \Rightarrow \log y = - \log x + \log c\]
On rearranging, we get,
\[ \Rightarrow \log y + \log x = \log c\]
We know that$\log a + \log b = \log \left( {ab} \right)$,
\[ \Rightarrow \log xy = \log c\]
On taking antilog on both sides we get,
$ \Rightarrow xy = c$
Where c is a constant.
This is an equation of a rectangular hyperbola. So, the required curve is a hyperbola.
Therefore the correct answer is option C.

Note: A hyperbola is one of the conic sections. It is formed by the intersection of a double right circular cone with a plane that passes through both the halves of the cone. It is the set of all the points in a plane which has a constant difference between distances from two fixed points. The fixed points are called foci of the hyperbola. A rectangular hyperbola is a hyperbola in semi-major and semi-minor axes are equal. We only need to apply the midpoint formula to any one of the coordinates. While doing integration, we must put the constant of integration. We take the constant as a log for easy calculations.