If tangents drawn to the parabola $(x-1)^2=8(y+1)$ which are perpendicular to the variable line $y=px-2p^2-p-1,$ where $p$ is a parameter, then point of intersection of these tangents to the variable line lies on the curve, which is:
- A) $\;y+3=0$
- B) $\;y=3$
- C) $\;x=3$
- D) $\;x+3=0$
If two tangents are perpendicular to one line, does that mean that line is normal to the curve at two point of contacts?
For $x^2=4ay$, the equation of normal is $y=mx+2a+\dfrac a{m^2}$
Using that here, I get $(y+1)=m(x-1)+4+\dfrac4{m^2}$
Rearranging the line in the question, $(y+1)=p(x-1)-2p^2$
On comparing, I get, $4+\dfrac4{p^2}=-2p^2$
But LHS is positive and RHS is non-positive. So, looks like I have made a mistake somewhere.
How to approach this question?
To simplify, let's use translation of coordinate system as $X = x - 1, Y = y + 1$
Then the equation of parabola is $X^2 = 8 Y$ and the variable line is $Y = p X - 2p^2 \tag1$
Slope of a line perpendicular to the variable line is $ - \dfrac 1p$.
Taking derivative, slope of tangent to the parabola at point $(u, v)$ is $\dfrac{u}{4}$
As tangent is perpendicular to the variable line,
$ \displaystyle u = - \frac 4 p, v = \frac{2}{p^2}$
So equation of the tangent line is,
$ \displaystyle Y - \frac 2 {p^2} = - \frac 1 {p^2} (pX + 4) \tag2$
To find the point of intersection of the variable line and the tangent to the parabola, we solve $(1)$ and $(2)$ and while $X$ is a function of $p$, $Y = - 2$.
So the locus of intersection points is $y = -3$.