Question

A conic C passing through $P(1,2)$ is such that the slope of its tangent at any point on the conic is inversely proportional to the ordinate of that point and conic C passes through origin.

If a circle touches the conic C at the point $P(1,2)$ and passes through the focus of the conic then its radius is-

• A. $1$
• B. $\sqrt{2}$
• C. $2$
• D. $\sqrt{3}$
• E. $\sqrt{5}$

Answer 1

Answer: (B)

$\sqrt{2}$

The equation of the conic in differential form will be

$\frac{dy}{dx}=\frac{k}{y}$ where k is the constant of proportionality.

Hence

$y.dy=k.dx$
$y^2=2kx+C$

Now it passes through the origin, hence $C=0$.

Therefore the equation reduces to

$y^2=2kx$

Now it also passes through $(1,2)$

Hence the equation of the parabola is

$y^2=4x$

$Focus=(1,0)$

Let the equation of the circle be $(x-\alpha {)}^2+(y-\beta {)}^2=r^2$

Now it passes through $(1,0)$ and $(1,2)$

Hence

$(1-\alpha {)}^2+{\beta }^2=r^2$
$(1-\alpha {)}^2+(2-\beta {)}^2=r^2$

Subtracting equation ii from i, we get

$2\beta -2=0$
$\beta =1$

Hence the equation of the circle will be

$(x-\alpha {)}^2+(y-1{)}^2=r^2$

Now the slope of the tangent to the parabola and the circle at $(1,2)$ will be equal.

$y^2=4x$
$2y.y^{\prime }=4$
$y_{(1,2)}^{\prime }=\frac{2}{y}=1$ ...(a)

$(x-\alpha {)}^2+(y-1{)}^2=r^2$
$2(x-\alpha )+2(y-1).y^{\prime }=0$
$y^{\prime }=\frac{\alpha -x}{y-1}$

Now

$y_{1,2}^{\prime }=\frac{\alpha -1}{1}=1$ ...from (a)

Or

$\alpha =2$

Hence the equation of the circle is of the form

$(x-2{)}^2+(y-1{)}^2=r^2$

Hence center of the circle is at $C=(2,1)$

Thus

$PC=r$

$PC=\sqrt{(2-1{)}^2+(1-2{)}^2}$
$=\sqrt{2}$

Hence the radius of the circle is $\sqrt{2}$ units.