Circle with centre on positive y-axis touches parabola at P and Q

Question

**Question: ** A circle with unit radius has its centre on positive y-axis. If this circle touches the parabola y=2x^2 tangentially at the points P and Q, then the sum of their ordinates is-

**Answer: ** 15/4

**Attempt: **

I assumed tangents on the parabola as $$ y = tx - t^2/8 $$ on the points P(t1) and Q(t2). Now this is also a tangent for the circle $$x^2 + (y-k)^2 = 1$$. Putting x from the tangent to the equation of circle. The obtained equation in y should have D=0. For it has one root only. Now the obtained D=0 in t should have two roots t1 and t2. But i am not getting them as anything relevant as from the sum and product of roots.

Answer 1

Solve simultaneously $y=2x^2$ and $x^2+(y-k)^2=1$ in terms of $y$ and get $$y^2+y\left(\frac 12-2k\right)+k^2-1=0$$

This has double roots, so the discriminant is zero. This leads to $k=\frac{17}{8}$ and the value of the $y$ coordinate is $$\frac{2k-\frac 12}{2}$$

By symmetry, the ordinates of $P$ and $Q$ are both equal to this, giving the required result $$\frac{15}{4}$$

Answer 2

$$x^2+(y-k)^2=1$$ and $$y=2x^2$$ give

$$4x^4+(1-4k)x^2+k^2-1=0$$

the discriminant should be zero.

$$\Delta=16k^2-8k+1-16k^2+16=0$$ thus $$k=\frac {17}{8} $$ and the root is then

$$x^2=\frac {4k-1}{8}=\frac {15}{16} $$

and $$y_1=y_2=2x^2=\frac {15}{8} $$

finally the sum is $$y_1+y_2=\frac {15}{4} $$