Answer
Verified
425.4k+ views
Hint: Assume the equation of the general tangents to both the parabolas given in the equation. Then simply equate the product of their slopes to , and make the point whose locus needs to be found out, (let’s call it ) satisfy the equations of tangents you assumed.
Let’s assume two parabola
\[{y^2} = 4ax\] ……………… (1)
And
\[{x^2} = 4ay\]. ……………….. (2)
As lines are touching the parabola, therefore lines are tangents on the parabolas.
We are going to write the equation of tangents for both parabolas. As the question says that the lines intersect each other normally at a point. We have to write the equation in slope form.
For \[{y^2} = 4ax\]
Let slope of the tangent \[ = {m_1}\]
Therefore; the slope of tangent = slope of the parabola
\[\therefore \dfrac{{dy}}{{dx}} = {m_1}\] ……………. (A)
\[2y\dfrac{{dy}}{{dx}} = 4a\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y}\] ………… (B)
Equating (A) and (B)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} = {m_1}\]
\[ \Rightarrow y = \dfrac{{2a}}{{{m_1}}}\] Put this value in equation (1)
Form equation (1)
\[{\left( {\dfrac{{2a}}{{{m_1}}}} \right)^2} = 4ax\]
\[ \Rightarrow x = \dfrac{a}{{m_1^2}}\]
Therefore; co-ordinate of point of the tangent is
\[x = \dfrac{a}{{m_1^2}}\] And \[ \Rightarrow y = \dfrac{{2a}}{{{m_1}}}\]
Equation of tangent is
\[\left( {y - \dfrac{{2a}}{{{m_1}}}} \right) = {m_1}\left( {x - \dfrac{a}{{m_1^2}}} \right)\]
\[ \Rightarrow y = mx + \dfrac{a}{{{m_1}}}\] …….. (C) Equation of tangent of parabola\[{y^2} = 4ax\].
Equation of tangent for parabola \[{x^2} = 4ay\]
Just replace \[x \to y\,,\,\,y \to x\,,\,{m_1} \to {m_2}\,and \to b\]in equation (C)
\[ \Rightarrow x = {m_2}y + \dfrac{b}{{{m_2}}}\] …….. (D)
\[ \Rightarrow x{m_2} = m_{_2}^2y + b\]
\[ \Rightarrow y = \dfrac{x}{{{m_2}}} - \dfrac{b}{{m_{_2}^2}}\] ……….. (E)
Compare this equation (E) with the general equation of a straight line\[y = mx + c\].
Slope of tangent \[ = \dfrac{1}{{{m_2}}}\]
As tangents are intersecting perpendicular to each other.
Therefore, the product of slope\[ = - 1\].
\[{m_1} \cdot \dfrac{1}{{{m_2}}} = - 1\]
\[ \Rightarrow {m_2} = - {m_1}\] ……………… (3)
Assume tangents are intersecting with each other at a point\[P(h,k)\].
Therefore, equation (C) and (E) must satisfy the point ‘P’.
Put the value of ‘P’ in equations (C) and (E).
From equation (C)
\[ \Rightarrow y = {m_1}x + \dfrac{a}{{{m_1}}}\]
\[ \Rightarrow k = {m_1}h + \dfrac{a}{{{m_1}}}\]
\[ \Rightarrow m_1^2h - {m_1}k + a = 0\] ……………. (5)
From equation (E)
\[ \Rightarrow y = \dfrac{x}{{{m_2}}} - \dfrac{b}{{m_{_2}^2}}\]
\[ \Rightarrow k = \dfrac{h}{{{m_2}}} - \dfrac{b}{{m_{_2}^2}}\]
\[ \Rightarrow m_{_2}^2k - {m_2}h + b = 0\] …………….(6)
Put the value \[{m_2} = - {m_1}\] in equation (6)
From equation (6) and (3)
\[ \Rightarrow {\left( { - {m_1}} \right)^2}k - \left( { - {m_1}} \right)h + b = 0\]
\[ \Rightarrow m_{_1}^2k + {m_1}h + b = 0\] …………….. (7)
Apply the cross-multiplication method in equation (5) and (7)
\[m_1^2h - {m_1}k + a = 0\] ……………. (5)
\[m_{_1}^2k + {m_1}h + b = 0\] …………….. (7)
\[ \Rightarrow \dfrac{{m_1^2}}{{ - kb - ha}} = \dfrac{{( - ){m_1}}}{{hb - ak}} = \dfrac{1}{{{h^2} - ( - k)k}}\]
From above we can say that
\[ \Rightarrow \dfrac{{m_1^2}}{{ - kb - ha}} = \dfrac{1}{{{h^2} - ( - k)k}}\]
\[ \Rightarrow m_1^2 = \dfrac{{ - kb - ha}}{{{h^2} + {k^2}}}\] …………….. (8)
Again from above
\[ \Rightarrow \dfrac{{( - ){m_1}}}{{hb - ak}} = \dfrac{1}{{{h^2} + {k^2}}}\]
\[ \Rightarrow ( - ){m_1} = \dfrac{{hb - ak}}{{{h^2} + {k^2}}}\]
Squaring both sides
\[ \Rightarrow m_{_1}^2 = \left( {\dfrac{{hb - ak}}{{{h^2} + {k^2}}}} \right)\] ……… (9)
Comparing equation (8) and (9)
From equation (8) and (9)
\[ \Rightarrow m_1^2 = \dfrac{{ - kb - ha}}{{{h^2} + {k^2}}} = {\left( {\dfrac{{hb - ak}}{{{h^2} + {k^2}}}} \right)^2}\]
\[ \Rightarrow - (kb + ha) = \dfrac{{{{\left( {hb - ak} \right)}^2}}}{{{h^2} + {k^2}}}\]
\[ \Rightarrow - (kb + ha)\left( {{h^2} + {k^2}} \right) = {\left( {hb - ak} \right)^2}\]
\[ \Rightarrow {\left( {hb - ak} \right)^2} + (kb + ha)\left( {{h^2} + {k^2}} \right) = 0\] …….(10)
Equation (10) is in terms of\[\left( {h,y} \right)\].
Substitute
\[\begin{gathered}
h \to x \\
k \to y \\
\end{gathered} \]
Locus of point of intersection in term of ‘x’ and ‘y’
\[ \Rightarrow {\left( {bx - ay} \right)^2} + (by + ax)\left( {{x^2} + {y^2}} \right) = 0\]
Note: Cross multiplication
Assume two equation be
\[{A_{1\;}}{x^2}{\text{ }} + {\text{ }}{B_1}{\text{x }} + {\text{ }}{C_{1\;}} = {\text{ }}0\], and
\[{A_2}{x^2}{\text{ }} + {\text{ }}{B_2}{\text{x }} + {\text{ }}{C_{2\;}} = {\text{ }}0\].
The coefficients of \[{x^2}\] are: \[{A_1}\] and\[{A_2}\].
The coefficients of \[x\] are: \[{B_1}\;and{\text{ }}{B_2}\].
The constant terms are: \[{C_1}\;and{\text{ }}\;{C_2}\]
To solve the equations in a simplified way,
\[\dfrac{{{x^2}}}{{{B_1}{C_2} - {B_2}{C_1}}} = \dfrac{x}{{{C_1}{A_2} - {C_2}{A_1}}} = \dfrac{1}{{{A_1}{C_2} - {A_2}{B_1}}}\]
Let’s assume two parabola
\[{y^2} = 4ax\] ……………… (1)
And
\[{x^2} = 4ay\]. ……………….. (2)
As lines are touching the parabola, therefore lines are tangents on the parabolas.
We are going to write the equation of tangents for both parabolas. As the question says that the lines intersect each other normally at a point. We have to write the equation in slope form.
For \[{y^2} = 4ax\]
Let slope of the tangent \[ = {m_1}\]
Therefore; the slope of tangent = slope of the parabola
\[\therefore \dfrac{{dy}}{{dx}} = {m_1}\] ……………. (A)
\[2y\dfrac{{dy}}{{dx}} = 4a\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y}\] ………… (B)
Equating (A) and (B)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} = {m_1}\]
\[ \Rightarrow y = \dfrac{{2a}}{{{m_1}}}\] Put this value in equation (1)
Form equation (1)
\[{\left( {\dfrac{{2a}}{{{m_1}}}} \right)^2} = 4ax\]
\[ \Rightarrow x = \dfrac{a}{{m_1^2}}\]
Therefore; co-ordinate of point of the tangent is
\[x = \dfrac{a}{{m_1^2}}\] And \[ \Rightarrow y = \dfrac{{2a}}{{{m_1}}}\]
Equation of tangent is
\[\left( {y - \dfrac{{2a}}{{{m_1}}}} \right) = {m_1}\left( {x - \dfrac{a}{{m_1^2}}} \right)\]
\[ \Rightarrow y = mx + \dfrac{a}{{{m_1}}}\] …….. (C) Equation of tangent of parabola\[{y^2} = 4ax\].
Equation of tangent for parabola \[{x^2} = 4ay\]
Just replace \[x \to y\,,\,\,y \to x\,,\,{m_1} \to {m_2}\,and \to b\]in equation (C)
\[ \Rightarrow x = {m_2}y + \dfrac{b}{{{m_2}}}\] …….. (D)
\[ \Rightarrow x{m_2} = m_{_2}^2y + b\]
\[ \Rightarrow y = \dfrac{x}{{{m_2}}} - \dfrac{b}{{m_{_2}^2}}\] ……….. (E)
Compare this equation (E) with the general equation of a straight line\[y = mx + c\].
Slope of tangent \[ = \dfrac{1}{{{m_2}}}\]
As tangents are intersecting perpendicular to each other.
Therefore, the product of slope\[ = - 1\].
\[{m_1} \cdot \dfrac{1}{{{m_2}}} = - 1\]
\[ \Rightarrow {m_2} = - {m_1}\] ……………… (3)
Assume tangents are intersecting with each other at a point\[P(h,k)\].
Therefore, equation (C) and (E) must satisfy the point ‘P’.
Put the value of ‘P’ in equations (C) and (E).
From equation (C)
\[ \Rightarrow y = {m_1}x + \dfrac{a}{{{m_1}}}\]
\[ \Rightarrow k = {m_1}h + \dfrac{a}{{{m_1}}}\]
\[ \Rightarrow m_1^2h - {m_1}k + a = 0\] ……………. (5)
From equation (E)
\[ \Rightarrow y = \dfrac{x}{{{m_2}}} - \dfrac{b}{{m_{_2}^2}}\]
\[ \Rightarrow k = \dfrac{h}{{{m_2}}} - \dfrac{b}{{m_{_2}^2}}\]
\[ \Rightarrow m_{_2}^2k - {m_2}h + b = 0\] …………….(6)
Put the value \[{m_2} = - {m_1}\] in equation (6)
From equation (6) and (3)
\[ \Rightarrow {\left( { - {m_1}} \right)^2}k - \left( { - {m_1}} \right)h + b = 0\]
\[ \Rightarrow m_{_1}^2k + {m_1}h + b = 0\] …………….. (7)
Apply the cross-multiplication method in equation (5) and (7)
\[m_1^2h - {m_1}k + a = 0\] ……………. (5)
\[m_{_1}^2k + {m_1}h + b = 0\] …………….. (7)
\[ \Rightarrow \dfrac{{m_1^2}}{{ - kb - ha}} = \dfrac{{( - ){m_1}}}{{hb - ak}} = \dfrac{1}{{{h^2} - ( - k)k}}\]
From above we can say that
\[ \Rightarrow \dfrac{{m_1^2}}{{ - kb - ha}} = \dfrac{1}{{{h^2} - ( - k)k}}\]
\[ \Rightarrow m_1^2 = \dfrac{{ - kb - ha}}{{{h^2} + {k^2}}}\] …………….. (8)
Again from above
\[ \Rightarrow \dfrac{{( - ){m_1}}}{{hb - ak}} = \dfrac{1}{{{h^2} + {k^2}}}\]
\[ \Rightarrow ( - ){m_1} = \dfrac{{hb - ak}}{{{h^2} + {k^2}}}\]
Squaring both sides
\[ \Rightarrow m_{_1}^2 = \left( {\dfrac{{hb - ak}}{{{h^2} + {k^2}}}} \right)\] ……… (9)
Comparing equation (8) and (9)
From equation (8) and (9)
\[ \Rightarrow m_1^2 = \dfrac{{ - kb - ha}}{{{h^2} + {k^2}}} = {\left( {\dfrac{{hb - ak}}{{{h^2} + {k^2}}}} \right)^2}\]
\[ \Rightarrow - (kb + ha) = \dfrac{{{{\left( {hb - ak} \right)}^2}}}{{{h^2} + {k^2}}}\]
\[ \Rightarrow - (kb + ha)\left( {{h^2} + {k^2}} \right) = {\left( {hb - ak} \right)^2}\]
\[ \Rightarrow {\left( {hb - ak} \right)^2} + (kb + ha)\left( {{h^2} + {k^2}} \right) = 0\] …….(10)
Equation (10) is in terms of\[\left( {h,y} \right)\].
Substitute
\[\begin{gathered}
h \to x \\
k \to y \\
\end{gathered} \]
Locus of point of intersection in term of ‘x’ and ‘y’
\[ \Rightarrow {\left( {bx - ay} \right)^2} + (by + ax)\left( {{x^2} + {y^2}} \right) = 0\]
Note: Cross multiplication
Assume two equation be
\[{A_{1\;}}{x^2}{\text{ }} + {\text{ }}{B_1}{\text{x }} + {\text{ }}{C_{1\;}} = {\text{ }}0\], and
\[{A_2}{x^2}{\text{ }} + {\text{ }}{B_2}{\text{x }} + {\text{ }}{C_{2\;}} = {\text{ }}0\].
The coefficients of \[{x^2}\] are: \[{A_1}\] and\[{A_2}\].
The coefficients of \[x\] are: \[{B_1}\;and{\text{ }}{B_2}\].
The constant terms are: \[{C_1}\;and{\text{ }}\;{C_2}\]
To solve the equations in a simplified way,
\[\dfrac{{{x^2}}}{{{B_1}{C_2} - {B_2}{C_1}}} = \dfrac{x}{{{C_1}{A_2} - {C_2}{A_1}}} = \dfrac{1}{{{A_1}{C_2} - {A_2}{B_1}}}\]
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
What organs are located on the left side of your body class 11 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE