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Let Q be $(at_2^2,2at_2)$

The distance PQ will be $$a^2(t_2^2-t_1^2)^2+4a^2(t_2-t_1)^2$$

I will replace $t_1$ with $t$

$$a^2(t^2-(t^2+\frac{4}{t^2}+4)^2+4a^2(t+t+\frac 2t)^2$$

$$16a^2(\frac{4}{t^2}+t^2+3)$$

Minimising it using derivatives

$$\frac{-8}{t^3}+2t=0$$ $$t=\pm 2\sqrt 2$$

The answer is $\pm \sqrt 2$

I checked my calculation a lot of time, but I keep ending up with the same thing

Another problem is to find $PQ$. Even if I do have the right value for $t$, I can’t seem to eliminate a. How can I solve it?

Quanto
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Aditya
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1 Answers1

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Note that $y'=\frac {2a}y$ and the tangent slope at P is $\frac1t$. The normal line PQ passes through $(at^2,2at)$ and its slope is $-\frac1{y'}=-t$. Then, its equation is,

$$y-2at=-t(x-at^2)$$

Substitute $y^2=4ax$ to get,

$$y^2 + \frac{4a}ty - 4a^2(2+t^2)=0$$

and,

$$y_1+y_2 = -\frac{4a}t,\>\>\>\>\> y_1y_2=- 4a^2(2+t^2)$$ $$(y_1-y_2)^2 = (y_1+y_2)^2 -4y_1y_2= 16a^2\left(\frac1{t^2}+2+t^2\right)$$

Evaluate,

$$PQ^2 = (y_1-y_2)^2 + (x_1-x_2)^2$$ $$= (y_1-y_2)^2+ \frac1{16a^2}(y_1^2-y_2^2)^2$$ $$= (y_1-y_2)^2\left(1+ \frac1{16a^2} (y_1+y_2)^2 \right)$$ $$=16a^2\left(\frac1{t^2}+2+t^2\right)\left(1+\frac1{t^2}\right)$$

Then, mnimize $PQ^2$ to obtain $t=\pm\sqrt2$.

Quanto
  • 97,352