Question on shifting of origins on coordinate axes

Question

I was going through some problems, and I found a question that I couldn’t solve. The question I am about to ask is a tiny part of a much bigger question. One of the steps involved is to find the tangent of the curve $y=x^2+6$ at the point (1,7)

$$x^2=4\frac 14(y-6)$$

The equation of tangent is $$Xx_1=2a(Y+y_1)$$

Here $X=x$ and $Y=y-6$

So $$x(1)=2\frac 14(y-6+7)$$ $$ x=\frac 12 (y+1)$$ $$2x=y+1$$

As it must be evident by now, this is not the correct tangent. Tangent calculated by another method is $2x=y-5$. What am I missing?

Answer 1

This result holds for the standard parabolas: The equation of tangent is $$Xx_1=2a(Y+y_1)$$ Now set Here $X=x$ and $Y+6=y$ then equation of parabola changes to standard form $X^2=Y$. Now the coordinates also change: $(1,7)$ changes to $(1,1)$. Now, apply your formula for $X^2=Y$ at point $(1,1)$. Applying the formula, we get $X(1)=2(\frac{1}{4})(Y+1)$ This will be your answer for the standard form, to get back to the original tangent equation replace with $X=x$ and $Y=y-6$ and you get $2x=(y-6)+1$ which is equal to $2x=y-5$.

Answer 2

Tangent has slope at given point the abscissa, ordinate are $x=1, y=7.$ Using this form straight line has equation:

$$ \dfrac{y-7}{x-1}= 2x|_{x=1}=2 $$

Simplify

$$ y= 2x+5. $$