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avatar+978 

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

Best Answer 

 #1
avatar+23575 
+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 \(m^2\).

 

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

 

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

 

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

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

 

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

 

laugh

 Nov 28, 2019
edited by heureka  Nov 28, 2019
 #1
avatar+23575 
+2
Best Answer

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 \(m^2\).

 

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

 

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

 

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

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

 

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

 

laugh

heureka Nov 28, 2019
edited by heureka  Nov 28, 2019
 #2
avatar+978 
+2

Clear explanation! Thanks!

CoolStuffYT  Nov 28, 2019
 #3
avatar+105476 
+1

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

 

 

cool cool cool

CPhill  Nov 28, 2019
 #4
avatar+23575 
+1

Thank you, CoolStuffYT!

Thank you, CPhill !

 

laugh

heureka  Nov 29, 2019

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