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# Tough Parabola Question

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437
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+1247

Here's the question

A line  $$y=mx+b$$ intersects the parabola $$y=x^2$$ at points $$A$$ and $$B$$ . The line $$AB$$ intersects the y-axis at the point $$P$$ . If  $$AP-BP=1$$ then find $$m^2$$.

I'm not sure how to approach it, I was thinking that I had to use distance formula but didn't know where to start.

Help would be greatly appreciated! Thanks!

Nov 27, 2019
edited by CoolStuffYT  Nov 27, 2019

#1
+25639
+2

A line $$y=mx+b$$  intersects the parabola $$y=x^2$$ at points $$A$$ and $$B$$.
The line $$AB$$ intersects the y-axis at the point  $$P$$.

If $$AP-BP=1$$  then find $$m^2$$.

$$\text{Let A=(x_A,y_A)=(x_A,x_{A}^2) } \\ \text{Let B=(x_B,y_B)=(x_B,x_{B}^2) } \\ \text{Let P=(x_P,y_P)=(0,y_P) }$$

$$\begin{array}{|l&rcll|} \hline BP: & m &=& \dfrac{y_P-y_B}{x_P-x_B} \\ & &=& \dfrac{y_P-x_{B}^2}{0-x_B} \\ &\mathbf{ m}&=& \mathbf{\dfrac{x_{B}^2-y_P}{x_B}} \\\\ AP: & m &=& \dfrac{y_A-y_P}{x_A-x_P} \\ & &=& \dfrac{x_{A}^2-y_P}{x_A-0} \\ &\mathbf{m} &=& \mathbf{\dfrac{x_{A}^2-y_P}{x_A}} \\ \hline \mathbf{ m} & = \mathbf{\dfrac{x_{B}^2-y_P}{x_B}}&=& \mathbf{\dfrac{x_{A}^2-y_P}{x_A}} \\ & x_A(x_{B}^2-y_P) &=& x_B(x_{A}^2-y_P)\\ & x_A x_{B}^2-x_Ay_P &=& x_B x_{A}^2-x_By_P \\ & y_P(x_B-x_A) &=& x_B x_{A}^2-x_A x_{B}^2 \\ & y_P(x_B-x_A) &=& -x_Ax_B(x_B-x_A) \\ (1) & \mathbf{y_P} &=& \mathbf{-x_Ax_B} \\ \hline &\mathbf{ m}&=& \mathbf{\dfrac{x_{B}^2-y_P}{x_B}} \quad | \quad \mathbf{y_P=-x_Ax_B} \\ & m &=& \dfrac{x_{B}^2-(-x_Ax_B)}{x_B} \\ & m &=& \dfrac{x_{B}^2+x_Ax_B}{x_B} \\ & m &=& \dfrac{x_{B}^2}{x_B}+\dfrac{x_Ax_B}{x_B} \\ (2) & \mathbf{m} &=& \mathbf{x_A+x_B} \\ \hline \end{array}$$

$$\begin{array}{|rclrcl|} \hline \mathbf{AP-BP} &=& {1} \qquad \text{or} \qquad BP = AP-1 \\\\ \mathbf{AB} &=& \mathbf{AP+BP} \\ AB &=& AP+AP-1 \\ \mathbf{AB} &=& \mathbf{2AP-1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline AB^2 &=& (x_A -x_B )^2+(y_A -y_B )^2 \\ &=& (x_A -x_B )^2+\left(x_{A}^2 -x_{B}^2 \right)^2 \\ &=& (x_A -x_B )^2+\Big( (x_{A} -x_{B}) (x_{A} +x_{B}) \Big)^2 \\ &=& (x_A -x_B )^2+ (x_{A} -x_{B})^2 (x_{A} +x_{B})^2 \\ &=& (x_A -x_B )^2 \Big(1 +( \underbrace{x_{A} +x_{B}}_{=m})^2\Big) \\ &=& (x_A -x_B )^2 (1 +m^2 ) \\ \mathbf{AB} &=& \mathbf{(x_A -x_B )\sqrt{1 +m^2} } \\\\ AP^2 &=& (x_A -x_P )^2+(y_A -y_P )^2 \\ &=& (x_A -0 )^2+(x_{A}^2 -y_P )^2 \\ &=& x_{A}^2+(x_{A}^2 -y_P )^2 \quad | \quad y_P=-x_Ax_B \\ &=& x_{A}^2+(x_{A}^2 + x_Ax_B )^2 \\ &=& x_{A}^2+\Big(x_A(\underbrace{x_A+x_B}_{=m}) \Big)^2 \\ &=& x_{A}^2+(x_A\times m)^2 \\ &=& x_{A}^2+x_{A}^2m^2 \\ &=& x_{A}^2(1+m^2) \\ \mathbf{AP} &=& \mathbf{x_A\sqrt{1 +m^2} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{AB} &=& \mathbf{2AP-1} \\\\ \mathbf{(x_A -x_B )\sqrt{1 +m^2} } &=& 2\times \mathbf{x_A\sqrt{1 +m^2} } -1 \\ \sqrt{1 +m^2}(x_A-x_B-2x_A) &=& -1 \\ \sqrt{1 +m^2}(2x_A-x_A+x_B) &=& 1 \\ \sqrt{1 +m^2}(\underbrace{x_A+x_B}_{=m}) &=& 1 \\ \sqrt{1 +m^2}\times m &=& 1 \quad \text{square both sides} \\ (1 +m^2)\times m^2 &=& 1 \\ (m^2)^2 + m^2 - 1 &=& 0 \\ m^2 &=& \dfrac{-1\pm \sqrt{1-4*(-1)} }{2} \\\\ m^2 &=& \dfrac{-1 {\color{red}+} \sqrt{1-4*(-1)} }{2} \quad | \quad m^2 > 0 ! \\ \mathbf{m^2} &=& \mathbf{\dfrac{-1 +\sqrt{5} }{2}} \\ \hline \end{array}$$

Nov 28, 2019
edited by heureka  Nov 28, 2019

#1
+25639
+2

A line $$y=mx+b$$  intersects the parabola $$y=x^2$$ at points $$A$$ and $$B$$.
The line $$AB$$ intersects the y-axis at the point  $$P$$.

If $$AP-BP=1$$  then find $$m^2$$.

$$\text{Let A=(x_A,y_A)=(x_A,x_{A}^2) } \\ \text{Let B=(x_B,y_B)=(x_B,x_{B}^2) } \\ \text{Let P=(x_P,y_P)=(0,y_P) }$$

$$\begin{array}{|l&rcll|} \hline BP: & m &=& \dfrac{y_P-y_B}{x_P-x_B} \\ & &=& \dfrac{y_P-x_{B}^2}{0-x_B} \\ &\mathbf{ m}&=& \mathbf{\dfrac{x_{B}^2-y_P}{x_B}} \\\\ AP: & m &=& \dfrac{y_A-y_P}{x_A-x_P} \\ & &=& \dfrac{x_{A}^2-y_P}{x_A-0} \\ &\mathbf{m} &=& \mathbf{\dfrac{x_{A}^2-y_P}{x_A}} \\ \hline \mathbf{ m} & = \mathbf{\dfrac{x_{B}^2-y_P}{x_B}}&=& \mathbf{\dfrac{x_{A}^2-y_P}{x_A}} \\ & x_A(x_{B}^2-y_P) &=& x_B(x_{A}^2-y_P)\\ & x_A x_{B}^2-x_Ay_P &=& x_B x_{A}^2-x_By_P \\ & y_P(x_B-x_A) &=& x_B x_{A}^2-x_A x_{B}^2 \\ & y_P(x_B-x_A) &=& -x_Ax_B(x_B-x_A) \\ (1) & \mathbf{y_P} &=& \mathbf{-x_Ax_B} \\ \hline &\mathbf{ m}&=& \mathbf{\dfrac{x_{B}^2-y_P}{x_B}} \quad | \quad \mathbf{y_P=-x_Ax_B} \\ & m &=& \dfrac{x_{B}^2-(-x_Ax_B)}{x_B} \\ & m &=& \dfrac{x_{B}^2+x_Ax_B}{x_B} \\ & m &=& \dfrac{x_{B}^2}{x_B}+\dfrac{x_Ax_B}{x_B} \\ (2) & \mathbf{m} &=& \mathbf{x_A+x_B} \\ \hline \end{array}$$

$$\begin{array}{|rclrcl|} \hline \mathbf{AP-BP} &=& {1} \qquad \text{or} \qquad BP = AP-1 \\\\ \mathbf{AB} &=& \mathbf{AP+BP} \\ AB &=& AP+AP-1 \\ \mathbf{AB} &=& \mathbf{2AP-1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline AB^2 &=& (x_A -x_B )^2+(y_A -y_B )^2 \\ &=& (x_A -x_B )^2+\left(x_{A}^2 -x_{B}^2 \right)^2 \\ &=& (x_A -x_B )^2+\Big( (x_{A} -x_{B}) (x_{A} +x_{B}) \Big)^2 \\ &=& (x_A -x_B )^2+ (x_{A} -x_{B})^2 (x_{A} +x_{B})^2 \\ &=& (x_A -x_B )^2 \Big(1 +( \underbrace{x_{A} +x_{B}}_{=m})^2\Big) \\ &=& (x_A -x_B )^2 (1 +m^2 ) \\ \mathbf{AB} &=& \mathbf{(x_A -x_B )\sqrt{1 +m^2} } \\\\ AP^2 &=& (x_A -x_P )^2+(y_A -y_P )^2 \\ &=& (x_A -0 )^2+(x_{A}^2 -y_P )^2 \\ &=& x_{A}^2+(x_{A}^2 -y_P )^2 \quad | \quad y_P=-x_Ax_B \\ &=& x_{A}^2+(x_{A}^2 + x_Ax_B )^2 \\ &=& x_{A}^2+\Big(x_A(\underbrace{x_A+x_B}_{=m}) \Big)^2 \\ &=& x_{A}^2+(x_A\times m)^2 \\ &=& x_{A}^2+x_{A}^2m^2 \\ &=& x_{A}^2(1+m^2) \\ \mathbf{AP} &=& \mathbf{x_A\sqrt{1 +m^2} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{AB} &=& \mathbf{2AP-1} \\\\ \mathbf{(x_A -x_B )\sqrt{1 +m^2} } &=& 2\times \mathbf{x_A\sqrt{1 +m^2} } -1 \\ \sqrt{1 +m^2}(x_A-x_B-2x_A) &=& -1 \\ \sqrt{1 +m^2}(2x_A-x_A+x_B) &=& 1 \\ \sqrt{1 +m^2}(\underbrace{x_A+x_B}_{=m}) &=& 1 \\ \sqrt{1 +m^2}\times m &=& 1 \quad \text{square both sides} \\ (1 +m^2)\times m^2 &=& 1 \\ (m^2)^2 + m^2 - 1 &=& 0 \\ m^2 &=& \dfrac{-1\pm \sqrt{1-4*(-1)} }{2} \\\\ m^2 &=& \dfrac{-1 {\color{red}+} \sqrt{1-4*(-1)} }{2} \quad | \quad m^2 > 0 ! \\ \mathbf{m^2} &=& \mathbf{\dfrac{-1 +\sqrt{5} }{2}} \\ \hline \end{array}$$

heureka Nov 28, 2019
edited by heureka  Nov 28, 2019
#2
+1247
-1

Clear explanation! Thanks!

CoolStuffYT  Nov 28, 2019
#3
+115955
+1

Thanks, heureka.....that WAS a tough one   !!!!

CPhill  Nov 28, 2019
#4
+25639
+1

Thank you, CoolStuffYT!

Thank you, CPhill !

heureka  Nov 29, 2019