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Equation of latusrectum

$$X-a=0$$$$x-2-2=0$$ $$x=4$$

Therefore $$(y+3)^2=8(4-2)$$ $$y=1,-7$$ Then point from which tangents are drawn are $(4,1)$ and $(4,-7)$.

Their point intersection will $(-4,-3)$, using the GM and AM property of point of intersection of tangent.

The answer is (0,-3). How is the right answer?

Aditya
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  • Do you know about pole-polar relationships? This point is the pole of the latus rectum. It also happens to be the intersection of the axis and directrix. – amd Feb 03 '20 at 21:52

1 Answers1

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Note that the tangent of the parabola is,

$$y' = \frac4{y+3}$$

So, at the two ends of the latus rectum with $y=1,-7$, they are $y'= 1,-1$, respectively. Hence, the corresponding equations of the two tangent lines are,

$$y-1=x-4,\>\>\>\>\>\>y+7 = -1(x-4)$$

They intersect at the point $(0,-3)$.

Quanto
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  • What is $y’$ and why is $y’=\frac{4}{y+3}$? – Aditya Feb 04 '20 at 11:55
  • @Aditya - take the derivative of $(y+3)^2=8(x-2)$ on both sides to get $2(y+3)y’=8$. Then, rearrange – Quanto Feb 04 '20 at 13:11
  • Okay, what’s wrong with my approach? All did was some fairly straightforward substitution – Aditya Feb 04 '20 at 14:20
  • @Aditya - Your approach is correct up to the point "Then point from which tangents are drawn are $(4,1)$ and $(4,-7)$". It's unclear from the post what you did afterwards – Quanto Feb 04 '20 at 14:28
  • I used a well-known property point of intersection of two tangents drawn to a parabola is (GM of abscissa, AM of ordinates) – Aditya Feb 04 '20 at 15:58
  • @Aditya - The way you used the property implicitly assumes that the $x$-coordinate of the vertex is 0, which is not the case here. – Quanto Feb 04 '20 at 16:34
  • Right. Then how should I modify the property to match the transformation, or is it just not possible? – Aditya Feb 04 '20 at 16:43
  • @Aditya - The generalized property states that the abscissa of the tangent intersection is $-(\frac p2 - x_0)$, where $p$ and $x_0$ are the $x$-coordinates of the focus and vertex, respectively. In your case $p=4$ and $x_0=2$. – Quanto Feb 04 '20 at 16:49
  • I would like to note this down. What will be the case in for ordinates? – Aditya Feb 04 '20 at 17:04
  • @Aditya - The case for ordinates remains the same, i.e. still the average of ordinates – Quanto Feb 04 '20 at 17:07
  • What if it’s the form of $(x-h)^2=4a(y-k)^2$ ie. pointing upwards rather than rightwards? Will y change then? – Aditya Feb 04 '20 at 17:18
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    @Aditya - Then, you just switch the coordinates for the intersection, i.e. its $y$-coordinate become $-(\frac p2 - y_0)$. – Quanto Feb 04 '20 at 17:39