Question

Through the focus of the parabola $x^{2}-4x-8y+44=0$, if tangents are drawn to another parabola $y^{2}=20x$, then the sum of the Y - coordinates of the points of contact of these tangents is

• A. 20
• B. 8
• C. 14
• D. 32

Hint:

The Y-coordinate of the point of contact on $y^2=4ax$ for a tangent line of slope $m$ is simply $\frac{2a}{m}$.

Answer 1

The Correct Option is C

Step 1: Find the focus of the first parabola.
Rewrite $x^2-4x-8y+44=0$ as $(x-2)^2=8(y-5)$. Here $4a=8$, so $a=2$, and the focus is $(2,5+2)=(2,7)$.
Step 2: Set the second parabola.
The second is $y^2=20x$, so $4a'=20$ and $a'=5$. We draw tangents from $(2,7)$ to it.
Step 3: Tangent with slope $m$.
A tangent is $y=mx+\dfrac{5}{m}$. It passes through $(2,7)$, so $7=2m+\dfrac{5}{m}$.
Step 4: Solve the slope equation.
Multiply by $m$: $2m^2-7m+5=0$, giving $m=1$ or $m=\dfrac52$.
Step 5: Y-coordinate of each contact point.
The point of contact on $y^2=4a'x$ has $y=\dfrac{2a'}{m}=\dfrac{10}{m}$. So the two Y-values are $\dfrac{10}{1}=10$ and $\dfrac{10}{5/2}=4$.
Step 6: Add the Y-coordinates.
\[ 10+4=14. \] \[ \boxed{14} \]