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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?

aarbee
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1 Answers1

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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 $y = -3$.

Math Lover
  • 51,819
  • Looks like I had completely misunderstood the question. Thanks for clarifying. – aarbee Feb 13 '22 at 07:01
  • Hi. Trying to solve $(1),(2)$. Equating $Y$ from both equations. $$pX-2p^2-\frac2{p^2}=-\frac Xp-\frac4{p^2}\\implies(p+\frac1p)X=2(p^2-\frac1{p^2})$$ How to proceed next? – aarbee Feb 13 '22 at 07:15
  • you should get $(p + 1/p) X = 2 (p^2 - 1/p^2)$ and you can simplify and that gives you $X$. But I did not find $X$. I only solved for $Y$. Note that From $(1)$, $pX = Y + 2p^2$. Plug that into $(2)$ – Math Lover Feb 13 '22 at 07:21
  • Got it, thanks. – aarbee Feb 13 '22 at 07:24