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Question: Find the curve for which the portion of the tangent included between the coordinate axes is bisected...

Find the curve for 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

Explanation

Solution

Here let us take point P(x,y)P(x,y) as the point of contact of tangent and curve. The slope of tangent is dydx\dfrac{dy}{dx} and using slope-point form we get the equation of tangent. After that find the points at which tangent meet both axes by substituting x=0x=0 and then y=0y=0. Then take the midpoint and integrate.

Complete step-by-step answer:
Here let us take point P(x,y)P(x,y) as the point of contact of tangent and curve and also slope of tangent is dydx\dfrac{dy}{dx}.
The equation of line through (x0,y0)({{x}_{0}},{{y}_{0}}) and mm as slope is yy0=m(xx0)y-{{y}_{0}}=m(x-{{x}_{0}}).
Now for equation of tangent let us substitute slope and points, we get,
Yy=dydx(Xx)Y-y=\dfrac{dy}{dx}(X-x)
Now let us find the coordinates on both the axes, for that let us consider YY intercept, where X=0X=0.
Yy=dydx(0x)Y-y=\dfrac{dy}{dx}(0-x)
Now simplifying above we get,
Y=yxdydxY=y-x\dfrac{dy}{dx} …………. (1)
So, the point becomes A(0,yxdydx)A\left( 0,y-x\dfrac{dy}{dx} \right).
Now, for XX intercept, Y=0Y=0, we get,
0y=dydx(Xx)0-y=\dfrac{dy}{dx}(X-x)
Now simplifying above we get,
X=xydxdyX=x-y\dfrac{dx}{dy}
So, the point is B(xydxdy,0)B\left( x-y\dfrac{dx}{dy},0 \right).
Now in question it is mentioned that it is bisected at point of contact.
So now, taking midpoint we get,
For xx axis part,
x=0+xydxdy2x=\dfrac{0+x-y\dfrac{dx}{dy}}{2}
On simplifying we get,
x=ydxdyx=-y\dfrac{dx}{dy}
On separating variables we get,
dyy=dxx\dfrac{dy}{y}=-\dfrac{dx}{x}
No integrating above both sides we get,
dyy=dxx\int{\dfrac{dy}{y}}=-\int{\dfrac{dx}{x}}
We know that, 1ada=loga+logc\int{\dfrac{1}{a}da=\log a+\log c}.
So, we get,
logy=logx+logc\log y=-\log x+\log c
Simplifying we get,
logy+logx=logc\log y+\log x=\log c
Now we know that, loga+logb=log(ab)\log a+\log b=\log (ab)
So, using above property we get,
logxy=logc\log xy=\log c
Taking antilog on both sides we get,
xy=cxy=c
Where cc is constant.
We can see that it is a curve of rectangular hyperbola.
The curve is hyperbola.
The correct answer is option (C).

Note: A hyperbola is the locus of all those points in a plane such that the difference in their distances from two fixed points in the plane is a constant. Hyperbola is defined as an open curve having two branches which are mirror images to each other. It is two curves that are like infinite bows. In other words, the locus of a point moving in a plane in such a way that the ratio of its distance from a fixed point (focus) to that from a fixed line (directrix) is a constant greater than 1.