# Two lines are drawn at the right angle, one being a tangent to \[{y^2} = 4ax\] and another to \[{x^2} = 4by\]. Show that the locus of their point of intersection is the curve \[ \ {\left( {bx - ay} \right)^2} + (by + ax)\left( {{x^2} + {y^2}} \right) = 0\]

Answer

Verified

363.9k+ 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,

\[\dfrac{{{x^2}}}{{{B_1}{C_2} - {B_2}{C_1}}} = \dfrac{x}{{{C_1}{A_2} - {C_2}{A_1}}} = \dfrac{1}{{{A_1}{C_2} - {A_2}{B_1}}}\]

Last updated date: 25th Sep 2023

•

Total views: 363.9k

•

Views today: 8.63k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE